n-linear algebra of type ii, by w. b. vasantha kandasamy, florentin smarandache

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n-LINEAR ALGEBRA OF

W. B. Vasantha Kandasamy

e-mail: [email protected]: http://mat.iitm.ac.in/~wbv

www.vasantha.in

Florentin Smarandachee-mail: [email protected]



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Peer reviewers:

Professor Diego Lucio Rapoport

Departamento de Ciencias y Tecnologia

Universidad Nacional de Quilmes

Roque Saen Peña 180, Bernal, Buenos Aires, Argentina

Dr.S.Osman, Menofia University, Shebin Elkom, EgyptProf. Mircea Eugen Selariu,

Polytech University of Timisoara, Romania.

Copyright 2008 by InfoLearnQuest and authors

Cover Design and Layout by Kama Kandasamy

Many books can be downloaded from the following

Digital Library of Science:

http://www.gallup.unm.edu/~smarandache/eBooks-otherformats

ISBN-10: 1-59973-031-6

ISBN-13: 978-1-59973-031-8

EAN: 9781599730318



CONTENTS

Preface

Chapter One

n-VECTOR SPACES OF TYPE II AND THEPROPERTIES

1.1 n-fields

1.2 n-vector Spaces of Type II



FURTHER READING

INDEX

ABOUT THE AUTHORS



PREFACE

This book is a continuation of the book n-linear

I and its applications. Most of the properties th

derived or defined for n-linear algebra of type I i

in this new structure: n-linear algebra of ty

introduced in this book. In case of n-linear algebare in a position to define linear functionals whi

marked difference between the n-vector spaces o

However all the applications mentioned in n-lin

type I can be appropriately extended to n-linear

II. Another use of n-linear algebra (n-vector spac

that when this structure is used in coding theordifferent types of codes built over different finite

this is not possible in the case of n-vector sp

Finally in the case of n-vector spaces of type II w

eigen values from distinct fields; hence the



suggests over a hundred problems. It is important tha

should be well versed with not only linear algebra linear algebras of type I.

The authors deeply acknowledge the unflinching

Dr.K.Kandasamy, Meena and Kama.

W.B.VASANTHA K

FLORENTIN SMA



Chapter One

n-VECTOR SPACES OF TYPE II A

THEIR PROPERTIES

In this chapter we for the first time introduce t

vector space of type II. These n-vector spaces

different from the n-vector spaces of type I becau

spaces of type I are defined over a field F where

spaces of type II are defined over n-fields. S

enjoyed by n-vector spaces of type II cannot bevector spaces of type I. To this; we for the sake o

just recall the definition of n-fields in section o

spaces of type II are defined in section two and

properties are enumerated.



We illustrate this by the following example.

Example 1.1.1: Let F = R Z3 Z5 Z17 be a 4-fie

Now how to define the characteristic of any n-field, n

DEFINITION 1.1.2: Let F = F 1 F 2 … F n be a say F is a n-field of n-characteristic zero if each f

characteristic zero, 1 d i d n.

Example 1.1.2: Let F = F1 F2 F3 F4 F5 F

= Q( 2 ), F2 = Q( 7 ), F3 = Q( 3 , 5 ), F4 = Q(

Q( 3 , 19 ) and F6 = Q( 5 , 17 ); we see all th

F2, …, F6 are of characteristic zero thus F is a

characteristic 0.

Now we proceed on to define an n-field of finite char

DEFINITION 1.1.3: Let F = F 1 F 2 … F n (n t

field. If each of the fields F i is of finite characteris zero characteristic for i = 1, 2, …, n then we call F field of finite characteristic.

Example 1.1.3: Let F = F1 F2 F3 F4 = Z5 Z31, F is a 4-field of finite characteristic.

Note: It may so happen that in a n-field F = F1 F2

n t 2 some fields Fi are of characteristic zero and



Example 1.1.4: Let F = F1 F2 F3 F4 F5

F2 = Z7, F3 = Q( 7 ), F4 = Q( 3, 5 ) and F5 = Qthen F is a 5-field of mixed characteristic

characteristic two, F2 is a field of characteristic

are fields of characteristic zero.

Now we define the notion of n-subfields.

DEFINITION 1.1.5: Let F = F 1 F 2 … F n b

2). K = K 1 K 2 … K n is said to be a n-subf

K i is a proper subfield of F i , i = 1, 2, …, n and

K i if i z j, 1 d i, j d n.

We now give an example of an n-subfield.

Example 1.1.5: Let F = F1 F2 F3 F4 where

F1 = Q( 2, 3 ), F2 = Q( 7, 5 )

F3 =2

2

Z [x]

x x 1

§ ·

¨ ¸¨ ¸ © ¹and F4 =

11

2

Z [x

x x

§

¨ ¨ ©

be a 4-field. Take K = K 1 K 2 K 3 K 4 = (Q(

Z2 Z11 F = F1 F2 F3 F4. Clearly K

of F.

It may so happen for some n-field F, we see it hso we call such n-fields to be prime n-fields.

Example 1.1.6: Let F = F1 F2 F3 F4 = Z



field. If all the fields Fi in F are non prime i.e., ar

fields then we call F to be a non prime field. Now isemiprime field or semiprime n-field if we have an

m < n then we call it as a quasi m-subfield of F m < n

We illustrate this situation by the following example

Example 1.1.7: Let

F = Q Z7 > @2

2

Z x

x x 1

> @7

2

Z x

x 1

> 3

3

Z

x

be a 5-field of mixed characteristic; clearly F is a 5

field. For Q and Z7 are prime fields and the other threnon-prime; take K = K 1 K 2 K 3 K 4 K 5 = I Z7 Z3 F = F1 F2 F3 F4 F5. Clearly K i

subfield of F.

1.2 n-Vector Spaces of Type II

In this section we proceed on to define n-vector spa

II and give some basic properties about them.

DEFINITION 1.2.1: Let V = V 1 V 2 … V n wher

a distinct vector space defined over a distinct field Fi = 1, 2, …, n; i.e., V = V 1 V 2 … V n is defined

field F = F 1 F 2 … F n. Then we call V to be space of type II.



Example 1.2.2: Let F = F1 F2 F3 where

Q( 3 ) u Q( 3 ) a vector space of dimension 3

= Q( 7 ) [x] be the polynomial ring with co

Q( 7 ), F2 is a vector space of infinite dimensi

and

F3 =

a b

a,b,c,d Q( 5)c d

- ½ª º° °® ¾« »° °¬ ¼¯ ¿

F3 is a vector space of dimension 4 over Q( 5 )

F2 F3 is a 3-vector space over the 3-field Q(

Q( 5 ) of type II.

Thus we have seen two examples of n-vector sp Now we will proceed on to define the linea

elements of the n-vector space of type II. Any el

V1 V2 … Vn is a n-vector of the form (DDn) or (D1

D2 … Dn

) where Di or Di V

and each Di is itself a m row vector if dimension

We call S = S1 S2 … Sn where each Si is

of Vi for i = 1, 2, …, n as the n-set or n-subset

space V over the n-field F. Any element D = (DDn) = {

1

1 1 1

1 2 n, ,...,D D D } {

2

2 2 2

1 2 n, ,...,D D D

{n

n n n

1 2 n, ,...,D D D } of V = V1 V2 …

{i

i i i

1 2 n, ,...,D D D } Vi for i = 1, 2, …, n and i

jD

ni and 1 d i d n



is a linearly independent subset of V i over F i , this m

for each i, i = 1, 2, … , n. If even one of the subse subset S; say { 1 2, ,...,

j

j j j

k D D D } is not a linearly

subset of V j over F j , then the subset S of V is not

independent subset of V, 1 d j d n. If in the n-subset S subset is not linearly independent subset we say S is

dependent subset of V over F = F 1 F 2 … F n.

Now if the n-subset S = S 1 S 2 … S n V; S subset of V i , i = 1, 2, …, n is such that some of the sulinearly independent subsets over F j and some of th

of V i are linearly dependent subsets of S i over F i , then we call S to be a semi n-linearly independent n-

F or equivalently S is a semi n-linearly dependent nover F.

We illustrate all these situations by the following exa

Example 1.2.3: Let V = V1 V2 V3 = {Q( 3 ) u{Q( 7 )[x]5 | this contains only polynomials of degr

or equal to 5}

a ba,b,c,d Q( 2)

c d

- ½ª º° °® ¾« »

¬ ¼° °¯ ¿

is a 3-vector space over the 3-field F = Q( 3 )

Q( 2 ) .

Take

1 0 0-ª º ª°



T = {(1, 3) (0, 2) (5, 1)} {x3, 1, x2 + 1,

0 1 1 0,1 0 0 0

- ½ª º ª º° °® ¾« » « »° °¬ ¼ ¬ ¼¯ ¿

= T1 T2 T3 V1 V2 V3

is a proper 3-subset of V. Clearly T is only a se

3-subset of V over F or equivalently 3-semi depof V, over F for T1 and T2 are linearly dependen

and V2 respectively over the fields Q( 3

respectively and T3 is a linearly independent su

Q( 2 ) . Take

P = P1 P2 P3

= {(1, 2) (2, 5), (5, 4), (–1, 0)} {1, x

1 0 0 1 1 1 1 1, , ,

0 1 1 0 1 1 1 0

- ½ª º ª º ª º ª º® ¾« » « » « » « »

¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼¯ ¿

V1 V2 V3

is a 3-subset of V. Clearly P is a 3-dependent 3-s

F.

Now we have seen the notion of n-independ

dependent n-subset and semi n-dependent n-sub

We would proceed to define the notion of n-basi

DEFINITION 1.2.3: Let V = V 1 V 2 … V

space over the n-field F = F 1 F 2 … F n. A



Example 1.2.4: Let V = V1 V2 … V4 be a 4-vover the field F = Q( 2 ) Z7 Z5 Q( 3 ) w

Q( 2 ) u Q( 2 ), a vector space of dimension two o

V2 = Z7 u Z7 u Z7 a vector space of dimension 3 over

V3 = 5a b c a, b,c,d,e,f Zd e f

- ½ª º° °® ¾« »¬ ¼° °¯ ¿

a vector space of dimension 6 over Z5 and

V4 = Q( 3 )u Q( 3 ) u Q( 3 )u Q( 3 )

a vector space of dimension 4 over Q( 3 ). Clearly

3, 6, 4) dimension over F. Since the dimension offinite we see V is a finite dimensional 4-vector space

Now we give an example of an infinite dimensio

space over the n-field F.

Example 1.2.5: Let V = V1 V2 V3 be a 3-vector

the 3-field F = Q( 3 ) Z3 Q( 2 ) where V1 = Q

a vector space of infinite dimension over Q( 3 ), V2

Z3 is a vector space of dimension 3 over Z3 and

V3 = a b ca,b,c,d,e,f Q 2

d e f - ½ª º° °® ¾« »

¬ ¼° °¯ ¿

a vector space of dimension 6 over Q( 2 ). Clearl

t f i fi it di i F



= ^ `1

1 1 1

1 2, ,..., nD D D ^ `2

2 2 2

1 2, ,..., nD D D … ^in V, provided there exists n-scalars

C = ^ `1

1 1 1

1 2, ,..., nC C C ^ `2

2 2 2

1 2, ,..., nC C C … ^ such that

= ( 1 2 … n ) =

^ `1 1

1 1 1 1 1 1

1 1 2 2 ... n nC C C D D D

^

2 2 2 2

1 1 2 2 .. C CD D

^ `1 1 2 2 ... n n

n n n n n n

n nC C C D D D .

We just recall this for it may be useful in case o

of elements of a n-vector in V, V a n-vector sp

field F.

DEFINITION 1.2.5: Let S be a n-set of n-vecto

space V. The n-subspace spanned by the n-set S

S n , where S i V i , i = 1, 2, …, n is def

intersection W of all n-subspaces of V which con

When the n-set S is finite n-set of n-vectors,S = ^ `

1

1 1 1

1 2, ,..., nD D D ^ `2

2 2 2

1 2, ,..., nD D D … ^we shall simply call W the n-subspace spanned bS.

In view of this definition it is left as an exercise

prove the following theorem.

THEOREM 1.2.1: The n-subspace spanned by

subset S = S 1 S 2 … S n of a n-vector space



^ `1

1 1 1

1 2 ... k D D D ^ `2

2 2 2

1 2 ... k D D D

̂ `1 2 ... n

n n n

k D D D

wherei

j iS D for each i = 1, 2, …, n and i j k

by

11 2

1 1 1... k S S S 21 2

2 2 2... k S S S

1 2 ... nk

n n nS S S .

If W 1 , …, W k are n-subspaces of V = V 1 V 2 …

W i = 1 2 ...i i i

nW W W for i = 1, 2, …, n, then

W = 11 2

1 1 1...k

W W W 1 2

2 2 2... kW W W

1 2 ... nk

n n nW W W

is a n-subspace of V which contains each of the sub

W 1 W 2 … W n.

THEOREM 1.2.2: Let V = V 1 V 2 … V n be

space over the n-field F = F 1 F 2 … F n. Let V

by a finite n-set of n-vectors ^ `1

1 1 1

1 2, , ..., m E E E ^ 2

1 E

… ^ `1 2, , ...,n

n n n

m E E E . Then any independent set

in V is finite and contains no more than (m1 , melements.

Proof: To prove the theorem it is sufficient if we

f h V f V i h i



=in

i i

j j

j 1

x

D

¦

=i in m

i i i

j kj k

j 1 k 1

x A

E¦ ¦

=i in m

i i i

kj j k

j 1 k 1

(A x )

E¦¦

=i im n

i i i

kj j k

k 1 j 1

(A x )

E¦¦ .

Since ni > mi this imply that there exists scalars

all zero such thatim

i i

kj j

j 1

A x 0

¦ ; 1 k mi.

Hencei i

i i i i i i

1 1 2 2 n nx x ... x 0D D D . This show

linearly dependent set. This is true for each i. H

holds good for S = S1 S2 … Sn.

The reader is expected to prove the following the

THEOREM 1.2.3: If V is a finite dimensional n-vethe n-field F, then any two n-basis of V have the

n-elements.

THEOREM 1.2.4: Let V = V 1 V 2 … V

space over the n-field F = F 1 F 2 … F n(n1 n2 n ) then



spanned by S. Then the n-subset obtained by adjoin

n-linearly independent.

THEOREM 1.2.6: Let W = W 1 W 2 … W n be a

of a finite dimensional n-vector space V = V 1 V 2

every n-linearly independent n-subset of W is finite aof a (finite) n-basis for W.

The proof of the following corollaries is left as an

the reader.

COROLLARY 1.2.1: If W = W 1 W 2 … W n is subspace of a finite dimensional n-vector space V

finite dimensional and dim W < dim V; i.e. (m1 , m(n1 , n2 , …, nn ) each mi < ni for i = 1, 2, …, n.

COROLLARY 1.2.2: In a finite dimensional n-vecto

V 1 V 2 … V n every non-empty n-linearly inde

of n-vectors is part of a n-basis.

COROLLARY 1.2.3: Let A = A1 A2 … An , be

space where Ai is a ni u ni matrix over the field F i athe n-row vectors of A form a n-linearly independe

vectors in 1 2

1 2 ... nnn n

n F F F . Then A is n-invertib

Ai is invertible, i = 1, 2, …, n.

THEOREM 1.2.7: Let W 1 = 1 1 1

1 2 ... nW W W 2 2 2

1 2 ... nW W W be finite dimensional n-subspa

vector space V then



W1 W2 = 1 2 1 2

1 1 2 2W W W W ... W

has a finite n-basis.

{1

1 1 1

1 2 k , ,...,D D D } {2

2 2 2

1 2 k , ,...,D D D ) … { D

which is part of a n-basis.

{1 1 1

1 1 1 1 1

1 2 k 1 n k , ,..., , ,...,

D D D E E } {

2

2 2 2

1 2 k, ,..., ,D D D E

{n n n

n n n n n n

1 2 k 1 2 n k, ,..., , , ,..., D D D E E E

is a n-basis of W1 and

{1 1 1

1 1 1 1 1 1

1 2 k 1 2 m k , ,..., , , ,..., D D D J J J }

{ 2 2 22 2 2 2 2 21 2 k 1 2 m k , ,..., , , ,..., D D D J J J } …

{n n n

n n n n n n

1 2 k 1 2 m k, ,..., , , ,..., D D D J J J }

is a n-basis of W2. Then the subspace W1 + W2 is

the n-vectors

{1 1 1

1 1 1 1 n 1 2

1 k 1 2 n k 1 2, , , , , , , , , , D D E E E J J J! ! !

{2 2 2

2 2 2 2 2 2 2 2

1 2 k 1 2 n k 1 2, , , , , , , , , , , D D D E E E J J J! ! !

{n n n

n n n n n n n n

1 2 k 1 2 n k 1 2, , , , , , , , , , D D D E E E J J ! !

and these n-vectors form an n-independent n-set

k k k k k k

i i j j rr x y z 0D E J ¦ ¦ ¦

t f k 1 2



for certain scalars k

k k k

1 2 nC ,C ,...,C true for k = 1, 2, …,But the n-set {

i i

i i i i i i

1 2 k 1 2 n, , , , , , ,D D D J J J! ! }, i = 1, 2

independent each of the scalars k

r z 0 true for each

n. Thus k k k k

i i j jx y 0D E ¦ ¦ for k = 1, 2, …, n

{k k k

k k k k k k

1 2 k 1 2 n k , , , , , , ,

D D D E E E! ! } is also an inde

each k

ix 0 and each k

jy 0 for k = 1, 2, …, n. Thu

{1 1 1 1

1 1 1 1 1 1 1 1 1

1 2 k 1 2 n k 1 2 m, , , , , , , , , , , D D D E E E J J J! ! !

{2 2 2 2 2

2 2 2 2 2 2 2 2 2

1 2 k 1 2 n k 1 2 m k, , , , , , , , , , , D D D E E E J J J! ! !

{n n n

n n n n n n n n

1 2 k 1 2 n k 1 2, , , , , , , , , , , D D D E E E J J J! ! !

is a n-basis for

W1 + W2 = { 1 2

1 1W W } { 1 2

2 2W W } … { W

Finally

dim W1 + dim W2 = {k 1 + m1 + k 1 + n1, k 2 + m2 + k 2 + n2, …, k n + mn

= {k 1 + (m1 + k 1 + n1), k 2 + (m2 + k 2 + n2), …, k n +

nn)}

= dim(W1W2) + dim (W1 + W2).

Suppose V = V1 V2 … Vn be a (n1, n2, …

dimensional n-vector space over F1 F2 … Fn.

B = {1

1 1 1

1 2 n, ,...,D D D } {2

2 2 2

1 2 n, ,...,D D D }

{ n n n }



Let {1

1 1 1

1 2 nx ,x , , x! } {2

2 2 2

1 2 nx , x , ,x! } …

be the n-coordinates of a given n-vector in theC.

= {1 1

1 1 1 1

1 1 n nx xE E! } {2

2 2 2

1 1 nx xE !

{n n

n n n n

1 1 n nx xE E! }

=1n

1 1

j j

j 1

x

E¦ 2n

2 2

j j

j 1

x

E¦ … nn

n

j

j 1

x

E¦

=1 1n n

1 1 i

j ij i

j 1 i 1

x P

D¦ ¦ 2 2n n

2 2 2

j ij i

j 1 i 1

x P

D¦ ¦ … nn

j 1

x

¦

=1 1n n

1 1 1

ij j i

j 1 i 1

(P x )

D¦¦ 2 2n n

2 2 2

ij j i

j 1 i 1

(P x )

D¦¦ … nn

j 1 ¦¦

Thus we obtain the relation

=1 1

n n1 1 1

ij j i

j 1 i 1

(P x )

D¦¦ 2 2

n n2 2 2

ij j i

j 1 i 1

(P x )

D¦¦ … n

n

j 1 ¦

Since the n-coordinates (1

1 1 1

1 2 ny , y , , y! ) ( 2

1y

… (n

n n n

1 2 ny , y , , y! ) of the n-basis B are uniqu

it followsk n

k k k

i ij j

j 1

y (P x )

¦ ; 1 d i d nk . Let Pi be th

whose i, j entry is the scalar k

ijP and let X an



X = P-1Y

X

1

X2

… Xn

= (P

1

)

-1

Y

1

(P2

)

-1

Y

2

… (PThis can be put with some new notational convenien

[]B = P[]C

[]C = P-1[]B.

In view of the above statements and results, we havefollowing theorem.

THEOREM 1.2.8: Let V = V 1 V 2 … V n be a f…, nn ) dimensional n-vector space over the n-field and C be any two n-basis of V. Then there is

necessarily invertible n-matrix P = P 1 P 2 … (n1 u n1 ) (n2 u n2 ) … (nn u nn ) with entries fr

… F n i.e. entries of P i are from F i , i = 1, 2, …, n

[] B = P[]C and []C = P -1[] B

for every n-vector in V. The n-columns of P are giv

[ i

j E ] B; j = 1, 2, …, n and for each i; i = 1, 2, …, n.

Now we prove yet another interesting result.

THEOREM 1.2.9: Suppose P = P 1 P 2 … P n i

(n2 u n2 ) … (nn u nn ) n-invertible matrix ove

F = F 1 F 2 … F n , i.e. P i takes its entries from

h i 1 2 L V V V V b



Proof: Let V = V1 V2 … Vn be a n-vect

n2, …, nn) dimension over the n-field F = F1 Let B = {(1

1 1 1

1 2 n, ,...,D D D ) ( 2 2

1 2, ,...,D D

(n

n n n

1 2 n, ,...,D D D )} be a set of n-vectors i

{(1

1 1 1

1 2 n, ,...,E E E ) (2

2 2 2

1 2 n, ,...,E E E ) … ( n

1 ,E E

ordered n-basis of V for which (i) is valid, it

each Vi which has (i

i i i1 2 n, , ,E E E! ) as its b

ini i

j kj k

k 1

P

E D¦ – I, true for every i, i = 1, 2, …,n.

Now we need only show that the vectors i

jE

these equations I form a basis, true for each i, i =Q = P –1. Then

r r

i i i i i

j j j kj k

j j k

Q Q PE D¦ ¦ ¦

(true for every i = 1, 2, …, n)

= i i i

kj jr k

j k

P Q D¦¦

= i i i

kj jr k

k j

P Q§ ·

D¨ ¸© ¹

¦ ¦

= i

k D

true for each i = 1, 2, …, n. Thus the subspace

set { ii i i1 2 n, ,...,E E E } contain { i

i i i1 2 n, ,...,D D D } and h

this is true for each i. Thus

B = {1

1 1 1

1 2 n, ,...,D D D } {2

2 2 2

1 2 n, ,...,D D D }

{ n n n }



Now we proceed onto define the notion

transformation for n-vector spaces of type II.

DEFINITION 1.2.7: Let V = V 1 V 2 … V n be

space of type II over the n-field F = F 1 F 2 …

= W 1 W 2 … W n be a n-space over the same

F 1 F 2 … F n. A n-linear transformation of typ

function T = T 1 T 2 … T n from V into W( )i i i i i i

i i iT C C T T D E D E for all i , i in V i

scalars C i F i. This is true for each i; i = 1, 2, …, n.

T[C + ] = (T 1 T 2 … T n )

[C 11 + 1 ] [C 22 + 2 ] … [C

= T 1(C 11 + 1 ) T 2(C 22 + 2 ) T n(C nn + n )

for all i V i or 1 2 … n V and C i

C 2 … C n F 1 F 2 … F n. We say I = I 1

is the n-identity transformation of type II from V to V

I = I (1 2 … n )

= I 1(1 ) I 2(

2 ) … I n(n )

= 1 2 … n

for all 1

2

… n

V = V 1 V 2 … Vthe n-zero transformation 0 = 0 0 … 0 of typ

into V is given by 0 = 01 02 … n

= 0 0 … 0, for all 1 2 … n V.



i=

ini i

j j

j 1

x

D¦

we have

Uii =

ini i

i j j

j 1

U x

D¦

=in

i i

j i j

j 1

x (U )

D

¦

=in

i i

j j

j 1

x

E¦

so that U is exactly the rule T which we define. Thi

= , i.e. if = 1 2 … n and = 1 2 …

i ii j jT D E , i j ni and i = 1, 2, …, n.

Now we proceed on to give a more explicit descripti

linear transformation. For this we first recall if V

dimensional vector space over Fi then Vi # in

iF =

if Wi is also a vector space of dimension mi over Ffield then Wi # im

iF .

Now let V = V1 V2 … Vn be a n-vector

the n-field F = F1 F2 … Fn of (n1, n2, …, nn

over F, i.e., each Vi is a vector space over Fi of d

over Fi for i = 1, 2, …, n. Thus Vi # Fi. Hence V =… Vn # 1 2 nn n n

1 2 nF F ... F . Similarly if W = W

… Wn is a n-vector space over the same n-field

m2, …, mn) is the dimension of W over F then W1



= {(1

1 1 1

1 2 nx ,x , , x! ) (2

2 2 2

1 2 nx ,x , , x!

( nn n n1 2 nx ,x , , x! )}.

T = (1 1

1 1 1 1

1 1 n nx xE E! ) (2

2 2 2

1 1 nx xE !

(n n

n n n n

1 1 n nx xE E! ).

If B = B1 B2 … Bn is a (n1 u n1, n2 u n2,

matrix which has n-row vector

(1

1 1 1

1 2 n, , ,E E E! ) (2

2 2 2

1 2 n, , ,E E E! ) … ( E

then T = B. In other words if i i i

k k1 k 2(B ,B ,E

2, …, n, then

T{(1

1 1 1

1 2 nx ,x , , x! ) … ( n n

1 2x ,x , ,!

= T1(1

1 1 1

1 2 nx ,x , , x! ) T2(2

2 2 2

1 2 nx ,x , , x! )

Tn(n

n n n

1 2 nx ,x , , x! ) =

1

1

1 1 1

1 111 1n

1 1 1

1 2 n

1 1

m 1 m n

B B

[x ,x ,..., x ]

B B

ª º« »« »« »¬ ¼

!

# #

!

2

2

2 2 2

2 2

11 1n

2 2 2

1 2 n

2 2

m 1 m n

B B[x ,x , , x ]

B B

ª º« »« »« »¬ ¼

!! # #

!



W = W1 W2 … Wn . Let R T =1 2

1 2

T TR R

the n-range of T = T1 T2 … Tn that is the vectors = (1 2 … n) in W = W1 W2

such that = T i.e., i i

j i jTE D for each i = 1, 2, …

some D in V. Leti

i i i

1 2 T, R E E and Ci Fi. There

i i

1 2 i, VD D such that i i

i 1 1TD E and i i

i 2 2T D E . Since

for each i; i i i

i 1 2T (C )D D = i i i

i 1 i 2C T TD D = i i

1C E

shows that i i i

1 2C E E is also ini

i

TR . Since this is true

we have

( 1 1 1

1 2C E E ) … ( n n n

1 2C E E ) R T =1 2

1 2

T TR R

Another interesting n-subspace associated with ttransformation T is the n-set N = N1 N2 … Nof the n-vectors V such that T = 0. It is a n-sub

because

(1) T(0) = 0 so N = N1 N2 … Nn is non emp

T = 0 then , E V = V1 V2 … Vn. i.e., … n and = 1 2 … n thenT(C + ) = CT + T

= C.0 + 0

= 0

= 0 0 … 0

so C + N = N1 N2 … Nn.

DEFINITION 1.2.8: Let V and W be two n-vector spa

same n-field F = F 1 F 2 … F n of dimension (n



THEOREM 1.2.11: Let V and W be n-vector

same n-field F = F 1 F 2 … F n and suppos(n1 , n2 , …, nn ) dimensional. T is a n-linear transV into W. Then n-rank T + n-nullity T = n-dim nn ).

Proof: Given V = V1 V

2 … V

nto be a

dimensional n-vector space over the n-field F =

Fn. W = W1 W2 … Wn is a n-vector

same n-field F. Let T be a n-linear transformati

W given by T = T1 T2 … Tn where Ti

linear transformation, and Vi is of dimension

vector spaces Vi and Wi are defined over Fi; i = 1n-rank T = rank T1 rank T2 … rank T

nullity T1 nullity T2 … nullity Tn. So

nullity T = dim V = (n1, n2, …, nn) i.e. rank Ti +

Vi = ni.

We shall prove the result for one Ti and it is true1, 2, …, n. Suppose {

i

i i i

1 2 k , , ,D D D! } is a basis

space of Ti. There are vectorsi i

i i i

k 1 k 2 n, , , D D D!

{i

i i i

1 2 n, , ,D D D! } is a basis of Vi, true for every i

We shall now prove that {i i

i i

i k 1 i nT ,...,T

D D } i

range of Ti. The vector i

i 1TD , i

i 2TD , …,i

i

i nT D ce

range of Ti and since i

i jT 0D for j d k

{ i iT T } h T h h



and accordingly the vector i =

i

i

n

i ir r

j k 1C

D¦ is in the n

Ti. Sincei

i i i

1 2 k , ,...,D D D form a basis for Ni, there mu

i

i i i

1 2 k , ,...,E E E in Fi such thatik

i i i

n r

r 1

D E D¦ . Thus

i i

i

k ni i i i

n r j j

r 1 j k 1

C

E D D¦ ¦ = 0

and sincei

i i i

1 2 n, ,...,D D D are linearly independent we

i i i

i i i i i

1 2 k k 1 nC C 0E E E ! ! . If r i is the ran

fact that i ii k 1 i nT , , TD D! form a basis for the range othat r i = ni – k i. Since k i is the nullity of Ti andimension of Vi we have rank Ti + nullity Ti = dim

see this above equality is true for every i, i = 1, 2

have rank Ti + nullity Ti = dimVi = ni; we see

equality is true for every i, i = 1, 2, …, n. We have

nullity T1) (rank T2 + nullity T2) … (rank Tn) = dim V = (n1, n2, …, nn).

Now we proceed on to introduce and study the al

linear transformations.

THEOREM 1.2.12: Let V and W be a n-vector spac same n-field, F = F 1 F 2 … F n. Let T and U transformation from V into W. The function (T + U

by (T + U) = T + U is a n-linear transformation



… Fn . Let U = U1 U2 … Un and T

Tn be two n-linear transformations from V inthe function (T + U) defined by (T + U)() =T

linear transformation from V into W.

For = 1 2 … n V = V1 V2

C1 C2 … Cn; consider

(T + U)(C + )= T(C + ) + U(C + )= CT() + T + CU() + U()

= CT() + CU() + T + U

= C(T + U) + (T + U)

which shows T + U is a n-linear transformatiogiven by

T + U = (T1 T2 … Tn) + (U1 U2

= (T1 + U1) (T2 + U2) … (T

since each (Ti + Ui) is a linear transformation f

true for each i, i = 1, 2, …, m; hence we see T +transformation from V into W.

Similarly CT is also a n-linear transformation

CT = (C1 C2 … Cn) (T1 T2 = C1T1 C2T2 … CnTn.

Now for (d + ) where d = d1 d2 … dn a

… Fn i.e., di Fi for i = 1, 2, …, n and =

n and = 1 2 … n consider



(we know from properties of linear transformation ea

linear transformation for i = 1, 2, …, n) so CT = C1T

… CnTn is a n-linear transformation as

CT(d + ) = C1T1(d11 + 1) C2T2(d

22 +

CnTn(dnn + n).

Hence the claim. CT is a n-linear transformation from

Now we prove about the properties enjo

collection of all n-linear transformation of V into W

W) denote the collection of all linear transformationW, to prove Ln (V, W) is a n-vector space over the

F1 F2 … Fn, where V and W are n-vector spa

over the same n-field F. Just now we have proved Lclosed under sum i.e. addition and also Ln (V, W

under the n-scalar from the n-field i.e. we have

defining (T + U)() = T + U for all = 1 2 V, T + U is again a n-linear transformation from V

Ln(V, W) is closed under addition. We have also

every n-scalar C = C

1

C

2

… C

n

F = F1 Fand T = T1 T2 … Tn; CT is also a n-linear tra

of V into W; i.e. for every T, U Ln (V,W), T + U

and for every C F = F1 F2 … Fn and for

Ln(V, W), CT Ln(V, W). Trivially 0 = 0 for everyserve as the n-zero transformation of V into W.

Thus Ln

(V, W) is a n-vector space over the sam= F1 F2 … Fn.

Now we study the dimension of Ln (V, W), wh

n-vector spaces are of finite dimension.



Proof: Given V = V1 V2 … Vn is a n-ve

the n-field F = F1 F2 … Fn of (n1, n2, …

over F. Also W = W1 W2 … Wn is a n-ve

the n-field F of (m1, m2, …, mn) dimension over

T2 … Tn be any n-linear transformation of t

o Wi, this is the only way the n-linear transform

is defined because Vi is defined over Fi and W

over Fi. Clearly W j (i j) is defined over F j ancannot imagine of defining any n-linear transfor

{1

1 1 1

1 2 n, , ,D D D! } {2

2 2 2

1 2 n, , ,D D D! } … {

be a n-ordered basis for V over the n-field F. W

basis if each basis of Vi is ordered and B1 = {1

1E

{ 2

2 2 2

1 2 m, , ,E E E! } … { n

n n n

1 2 m, , ,E E E! } be a of W over the n-field F.

For every pair of integers (pi, qi); 1 d pi d mi

i = 1, 2, …, n, we define a n-linear transfo1 1 2 2 n n p ,q p ,q p ,q

1 2 nE E E ! from V into W by

i i

i

i

p ,q i

i j i i

p

0 if j qE ( )

if j q

- z°D ®

E °̄= i i

i

jq pG E

true for i = 1, 2, …, ni, i = 1, 2, …, mi. We

theorems a unique n-linear transformation frosatisfying these conditions. The claim is

transformationsi i p ,q

iE form a basis of L (Vi, Wi)

each i So Ln (V W) = L (V1 W1) L (V2 W2)



i

i i

i

mi i i

i j p j p

p 1

T A

D E¦ .

We wish to show that T =i i

i i

i i

i i

m ni p ,q

i p q p 1 q 1

A E

¦¦Let U be a n-linear transformation in the right hand

(I) then for each ji i

i i

i i

i i p ,q i

i j i j p q p q

U A E ( )D D¦¦

= i i i i

i i

i i

p q jq p p q

A G E¦¦

=i

i i j

i

mi i

p p

p 1

A

E¦

=Tii

jD

and consequently Ui = Ti as this is true for each i, i =

We see T = U. Now I shows that the E p,q spans Ln

each

i i p ,q

iE spans L(Vi, Wi); i = 1, 2, …, n. Now itshow that they are n-linearly independent. But this i

what we did above for if in the n-transformation U, w

i i

i i

i i

i p ,q

i i p q p q

U A E ¦¦

to be the zero transformation then ii jU 0D for each

soi

i i j

i

mi i

p p p 1

A 0

E ¦ and the independence of the i

i

pE i



from W into Z. Then the n-composed n-functionUT() = U(T()) is a n-linear transformation fro

Proof: Given V = V1 V2 … Vn , W = W

Wn and Z = Z1 Z2 … Zn are n-vector sp

field F = F1 F2 … Fn where Vi, Wi an

spaces defined over the same field Fi, this is true

n. Let T = T1 T2 … Tn be a n-linear transV into W i.e. each Ti maps Vi into Wi for i = 1

other way there can be a n-linear transformation

V into W. U = U1 U2 … Un is a n-linear

of type II from W into Z such that for every

transformation from Wi into Zi; (for Wi and Zi a

spaces defined over the same field Fi); this is tru1, 2, …, n. Suppose the n-composed function

(UT)() = U(T()) where = 1

2 …

n

UT = U1T1 U2T2 … UnTn and

UT() = (U1T1 U2T2 … UnTn)(1 2

= (U1T1)(1

) (U2T2)(2

) … ( U

Now from results on linear transformations we

defined by (UiTi)(i) = Ui(Ti(

i) is a linear trans

Vi into Zi; This is true for each i, i = 1, 2, …, n

n-linear transformation of type II from V into Z.

We now define a n-linear operator of type II for

V over the n-field F.

1 2 9



LEMMA 1.2.1: Let V = V 1 V 2 … V n be a n-v

over the n-field F = F 1 F 2 … F n of type II. Le

T 2 be the n-linear operators on V and let C = C 1

C n be an element of F = F 1 F 2 … F n. Then

a. IU = UI = U where I = I 1 I 2 …

identity linear operator on V i.e. I i (vi ) = vi

V i ; I i; V i o V i , i = 1, 2, …, n.b. U(T 1 + T

2 ) = UT 1 + UT

2 (T

1 + T 2 )U = T

1U + T 2U c. C(UT 1 ) = (CU)T

1 = U(CT 1 ).

Proof: (a) Given V = V1 V2 … Vn, a n-v

defined over the n-field of type II for F = F1 F2 each Vi of V is defined over the field Fi of F. If I = I

In: V o V such that Ii (i) = i for every i Vi;

1, 2, …, n then I() = for every V1 V2 …

and UI = (U1 U2 … Un) (I1 … In) = U1

… UnIn. Since each Ui Ii = Ii Ui for i = 1, 2, …, n,

= I1U1 I2U2 … InUn. Hence (a) is proved.

[U(T1 + T2)]() = (U1 U2 … Un) [(1 1

1 2T T

( 2 2 2

1 2 nT T ... T )] [1

2

= (U1 U2 … Un)[(1 2

1 1T T ) ( 1

2T T

( 1 2n nT T )] (1 2 … n)

= U1(1 2

1 1T T )1U2(1 2

2 2T T )(2) … Un( T

Si f h i h



a vector space over the field Z2. Clearly V1 is on

space over Z2 and never a linear algebra over Z2.

polynomials in x with coefficients from Q}; V2

algebra over Q.

V3 = 7

a b

c d a,b,c,d,e,f Z

e f

- ½ª º° °« » ® ¾« »° °« »¬ ¼¯ ¿

;

V3 is a vector space over Z7 and not a linear algebra

only a 3-vector space over the 3-field F and not

algebra over F.

We as in case of finite n-vector spaces V and W

F of type II, one can associate with ever

transformation T of type II a n-matrix. We in case

space V over the n-field F for every n-operator on V

n-matrix which will always be a n-square mixed ma

obvious by taking W = V, then instead of getting a n-order (m1n2, m2n2, …, mnnn) we will have mi = ni f

= 1, 2, …, n. So the corresponding n-matrix would

n-square matrix of n-order ( 2 2 2

1 2 nn ,n ,...,n ) provided

(n1, n2, …, nn). But in case of n-linear transformatio

we would not be in a position to talk about inverti

transformation. But in case of n-linear operators we

invertible n-linear operators of a n-vector space ove

F.

DEFINITION 1.2.11: Let V = V 1 V 2 … V n and

W 2 … W n be two n-vector spaces defined over



The following theorem is immediate.

THEOREM 1.2.15: Let V = V 1 V 2 … V nW 2 … W n be two n-vector spaces over the n

F 2 … F n of type II. Let T = T 1 T 2 …

transformation of V into W of type II. If T is n-inv

n-inverse n-function

1 1 1

1 2 ... nT T T T

transformation from W onto V.

Proof: We know if T = T1 T2 … T

transformation from V into W and when T is onto then there is a uniquely determined n-in

1 1 1 1

1 2 nT T T ... T

which map W onto Vis the n-identity n-function on V and TT-1 is t

function on W. We need to prove only if a n-line

is n-invertible then the n-inverse T-1 is also n-1 2 n

1 1 1...E E E , 2 =1 2 n

2 2 2...E E E be two

and let C = C

1

C

2

… C

n

be a n-scalar ofF1 F2 … Fn. Consider

T-1(C1 + 2) = ( 1 1 1 1

1 2 nT T T ... T )

[ 1 1 1 2 2 2

1 2 1 2C C ...E E E E

= 1 1 1 1 1 2 2

1 1 1 2 2 1 1

T (C ) T (C E E E 1 n n n

n 1 1 2T (C ) E E .

Since from usual linear transformation pro



Thus i i i

1 2C D D is the unique vector in Vi which is

into Ti (

i i i

1 2C E E ) and so1 i i i

i 1 2T (C ) E E = i i i

1 2C D D = i 1 i 1

i 1 iC T T E E

and 1

iT is linear. This is true for every Ti i.e., for i,

n. So 1 1 1 1

1 2 nT T T ... T is n-linear. Hence th

We say a n-linear transformation T is n-non-sing

0 implies J = (0 0 … 0); i.e., each Ti in T is ni.e. TiJi = 0 implies Ji = 0 for every i; i.e., if J = J1 Jn then TJ = 0 … 0 implies J = 0 0 … 0.

Thus T is one to one if and only if T is n-non sin

The following theorem is proved for th

transformation from V into W of type II.

THEOREM 1.2.16: Let T = T 1 T 2 … T n be

transformation from V = V 1 V 2 … V n into W

… W n. Then T is n-non singular if and only each n-linearly independent n-subset of V into aindependent n-subset of W.

Proof: First suppose we assume T = T1 T2 … singular i.e. each Ti: Vi o Wi is non singular for i =

Let S = S1 S2 … Sn be a n-linearly, n-ind

subset of V. If {1

1 1 1

1 2 k , ,...,D D D } {2

2 2 2

1 2 k , ,...,D D D

{n

n n n1 2 k , ,...,D D D } are n-vectors in S = S1 S2 …

n-vectors {1

1 1 1

1 1 1 2 1 k T ,T ,...,TD D D } { 2 2

2 1 2 2 2T ,T ,...,TD D

{ n n n

n 1 n 2 n kT ,T ,...,TD D D } are n-linearly n-indepen



Suppose T = T1 T2 … Tn is such

independent n-subsets onto independent n-subs

2 … n be a non zero n-vector in V i.e.

zero vector of Vi, i = 1, 2, …, n. Then the set S

Sn consisting of the one vector = 1 2 independent each

iis independent for i = 1, 2

image of S is the n-set consisting of the one n-v

T22 … T

nn and this is n-independent, t

because the n-set consisting of the zero n-

dependent. This shows that the n-nullspace of T

Tn is the zero subspace as each Tii 0 imp

zero vector alone in Vi is dependent for i = 1,

each Ti is non singular so T is n-non singular.

We prove yet another nice theorem.

THEOREM 1.2.17: Let V = V 1 V 2 … V n an

… W n be n-vector spaces over the same n-fi

… F n of type II. If T is a n-linear transform

from V into W, the following are equivalent.

(i) T is n-invertible(ii) T is n-non-singular (iii) T is onto that is the n-range of T is W

Proof: Given V = V1 V2 … Vn is a n-vethe n-field F of (n1, n2, … , nn) dimension over t

dimension of Vi over the field Fi is ni for i = 1, further assume n-dim(V) = (n1, n2, … , nn) =



Clearly T is a n-linear transformation of type II,

maps V onto 1 2 nn n n

1 2 nF F ... F or each Ti is li

one and maps Vi to in

iF for every i, i = 1, 2, …,

as in case of vector spaces transformation by

representation of n-transformations by n-matrice

Let V = V1 V2 … Vn be a n-vector sp

field F = F1 F2 … Fn of (n1, n2, …, nn) dimLet W = W1 … Wn be a n-vector space o

field F of (m1, m2, …, mn) dimension ove

{1

1 1 1

1 2 n, , ,D D D! } {2

2 2 2

1 2 n, , ,D D D! } … {

be a n-basis for V and C= (1

1 1 1

1 2 m, , ,E E E! ) ( 2

1E

… (n

n n n1 2 m, , ,E E E! ) be an n-ordered basis for

n-linear transformation of type II from V intdetermined by its action on the vectors i. Each

…, nn) tuple vector; i

i jTD is uniquely express

combination;

i

i i

i

mi i i

i j k j k k 1

T AD E¦ . This is true for e

…, n and 1 j ni of i

i

k E ; the scalars i

ijA ,.

coordinates of Tii

jD in the ordered basis { i

1,E E

This is true for each i, i = 1, 2, …, n.

Accordingly the n-transformation T = T1 determined by the (m1n1, m2n2, …, mnnn) scalars

ni matrix Ai defined by iA = iA is called



Tii= (Ti

ini i

j j

j 1

x

D¦ )

=in

i i

j i j

j 1

x (T )

D¦

=i i

i i

n mi i i

j k j k

j 1 k 1

x A

E¦ ¦

=i i

i i

m n i i i

k j j k

k 1 j 1

(A x )

E¦¦ .

If Xi is the coordinate matrix of i in the component

then the above computation shows that Ai Xi is the

matrix of the vector i

iT D ; that is the component of th

because the scalar i

i i

ni i

k j k

j 1

A x

¦ is the entry in the k th

column matrix Ai Xi. This is true for every i, i = 1, 2

us also observe that if Ai is any mi u ni matrix over

theni i i

i i

n m ni i i i i

i j j k j j k

j 1 k 1 j 1

T ( x ) ( A x )

D E¦ ¦ ¦

defines a linear transformation Ti from Vi into Wi, th

which is Ai relative to

i

i i

1 n{ ,..., }D D and i i

1 2{ , ,...,E E E

true of every i. Hence T = T1 T2 … Tn is

transformation from V = V1 V2 … Vn into

W2 … Wn, the n-matrix of which is A = A1



B =1

1 1 1

1 2{ , ,..., }nD D D 2

2 2 2

1 2{ , ,..., }nD D D …

= B

1

B

2

… B

n

be the n-basis of V over F C =

1

1 1 1

1 2{ , ,..., }m E E E 2

2 2 2

1 2{ , ,..., }m E E E … {

= C 1 C 2 … C n ,

n-basis for W, where W is a n-vector space ov field F of (m1 , m2 , …, mn ) dimension over F. Fo

transformation T = T 1 T 2 … T n of type I

there is a n-matrix A = A1 A2 … An wher

u ni matrix with entries in F i such that > @C

T AD

every n-vector = 1 2 … n V we hav

11

1 ... nn

nC C T T D D ª º ª º ¬ ¼ ¬ ¼ = 1

1 1 ... B

A D ª º ¬ ¼

Further more T o A is a one to one corresponthe set of all n-linear transformation from V in

and the set of all (m1 u n1 , m2 u n2 , …, mn u nn )

the n-field F. The n-matrix A = A1

A2

…associated with T = T 1 T 2 … T n is calledT relative to the n-ordered basis B and C. From

2

1 2 ...i n

j J n jT T T TD D D D =

1 2

1 2

1 2

1 1 2 2

1 1 1

...

n

n

n

mm m

nk j j k j j k j

k k k

A A A E E

¦ ¦ ¦

says that A = A1 A2 … An is the n-m



[ ] [ ] [i i

i i i i i i i i i

j j i j i j i j i jC CC A P C T U C T U D D D D

true for every i, i = 1, 2, … , n and 1 j ni.

Several interesting results analogous to usual vector

be derived for n-vector spaces over n-field of typ

related n-linear transformation of type II.

The reader is expected to prove the following the

THEOREM 1.2.20: Let V = V 1 V 2 … V n be a

nn ), n-vector space over the n-field F = F 1 F 2

let W = W 1 W 2 … W n be a (m1 , m2 , …, mn ) d

vector space over the same n-field F = F 1 … F pair of ordered n-basis B and C for V and W respec

function which assigns to a n-linear transformation its n-matrix relative to B, C is an n-isomorphism bet

space Ln (V, W) and the space of all n-matrices of n-

n1 , m2 u n2 , … , mn u nn ) over the same n-field F.

Since we have the result to be true for every pai

spaces Vi and Wi over Fi, we can appropriately extenfor V = V1 V2 … Vn and W = W1 W2 …

F = F1 F2 … Fn as the result is true for every i

Yet another theorem of interest is left for the reader t

THEOREM 1.2.21: Let V = V 1 V 2 … V n , W =

… W n and Z = Z 1 Z 2 … Z n be three (n1 , n2 , m2 , …, mn ), and (p1 , p2 , …, pn ) dimensional n-ve

respectively defined over the n field F = F F



for the n-spaces V, W and Z respectively, if A =

An is a n-matrix relative to T to the pair B, C and

… Rn is a n-matrix of U relative to the pair Cn-matrix of the composition UT relative to the p

product n-matrix E = RA i.e., if E = E 1 E 2

R2 A2 … Rn An.

THEOREM 1.2.22: Let V = V 1 V

2 … V

nnn ) finite dimensional n-vector space over the n

F 2 … F n and let B = B1 B2 … Bn =

2

2 2 2

1 2{ , , , }nD D D ! … 1 2{ , , , }n

n n n

nD D D ! an

… C n =1

1 1 1

1 2{ , , , }m E E E ! 2 2

1 2{ , , E E !

1 2{ , , , }n

n n nm E E E ! be an n-ordered n-basis for V

n-linear operator on V.

If P = P 1 P 2 … P n =1

1 1

1{ ,..., } {n P P

1 2{ , ,..., }n

n n n

n P P P is a (n 1 u n1 , n2 u n2 , … , nn

with jth

component of the n-columns [ ]i i

j j B P E 1, 2, …, n then 1[ ] [ ]C BT P T P i.e. 11 2[ ] [

CT T

= 1 2

1 1 1

1 1 1 2 2 2[ ] [ ] ... [ ] nn n n B B B P T P P T P P T P . A

is the n-invertible operator on V defined by iU D

…, ni , i = 1, 2, …, n then1

[ ] [ ] [ ] [ ]C B B BT U B T U

1 21 2[ ] [ ] ... [ ] nnC C C

T T T =

1 1 1 2 2 2

1 1

1 1 1 2 2 2[ ] [ ] [ ] [ ] [ ] [ ] ... [ n B B B B B BU T U U T U U



square matrix of n-order (n1 u n1 , n2 u n2 , …, nn u n same n-field F. We say that B is n-similar to A over t

if there is an invertible n-matrix P = P 1 P 2 …order (n1 u n1 , n2 u n2 , …, nn u nn ) over the n-field F = P -1 A P i.e.

B1 B2 … Bn =1 1

1 1 1 2 2 2 ... n P A P P A P P

Now we proceed on to define the new notion functionals or linear n-functionals. We know we w

position to define n-linear functionals in case of n-v

over the field F i.e. for n-vector spaces of type I.

Only in case of n-vector spaces defined over th

type II we are in a position to define n-linear functio

n-vector space V.Let V = V1 V2 … Vn be a n-vector space

field F = F1 F2 … Fn of type II. A n-linear tran

f = f 1 f 2 … f n from V into the n-field of n-sc

called a n-linear functional on V i.e., if f is a n-funct

into F such that f(CD + E) = Cf(D) + f(E) where C =

… Cn

; i = 1, 2, …, n, = 1

2

… n

and … n where i, i Vi for each i, i = 1,2, …, n. f

… f n where each f i is a linear functional on Vi; i =i.e.,

f(C + ) = (f 1 f 2 … f n) (C11 + 1) (C

… (Cnn + n)= (C1f 1(

1) + f 1(1)) (C2f 2(

2) + f 2(

(Cnf n(n) + f n(

n))]

= f1(C11 + 1) f2(C

22+ 2) f



linear function f = f 1 f 2 … f n from 1n

1F F

F1 F2 … Fn given by f 1( 1

1 1

1 ,..., n x x ) f 2(f n(

n

n n

1 nx ,..., x ) = (1 1

1 1 1 1

1 1 n nx ... xD D ) ( 2 2

1 1xD

… (2 n

n n n n

1 1 n nx ... xD D ) where i

jD are in Fi, 1

i = 1, 2, … , n; is a n-linear functional on 1n

1F F is the n-linear functional which is represented

[1

1 1 11 2 n, ,...,D D D ] [

2

2 2 21 2 n, ,...,D D D ] …

relative to the standard ordered n-basis for 1n

1F

and the n-basis {1} {1} … {1} for F = F

Fn.i

jD =i

i jf (E ); j = 1, 2, …, ni for every i = 1,

n-linear functional on1 2 nn n n

1 2 nF F ... F is osome n-scalars (

1

1 1

1 nD D! ) (2

2 2

1 nD D! ) …

That is immediate from the definition of the n-l

of type II because we define i i

j i jf (E )D .

?f 1(1

1 11 nx , , x! ) f 2(

2

2 21 nx , , x! ) … f n( x

=in

1 1

1 j j

j 1

f x E

§ ·¨ ¸

© ¹¦

2n2 2

2 j j

j 1

f x E

§ ·¨ ¸

© ¹¦ … nf

§¨©

= 1 1

j 1 j

j

x f (e ) ¦ 2 2

j 2 j

j

x f (e ) ¦ …

j

x¦

=1 2 nn n n

1 1 2 2 n n

j j j j j j

j j 1 j 1

a x a x ... a x

¦ ¦ ¦



dim Vi for every i. Thus dim V = dim V* = dim *

1V

… dim

*

nV . If B = { 1

1 1 1

1 2 n, ,...,D D D } {

2 2

1 2, ,...,D D D{

n

n n n

1 2 n, ,...,D D D } is a n-basis for V, then we know fo

functional of type II. f = f 1 f 2 … f n we have

such that k k k

i j ijf ( )D G true for k = 1, 2, …, n. In t

obtain from B a set of n-tuple of ni sets of disti

functionals { 11 1 11 2 nf ,f ,..., f }{ 2

2 2 21 2 nf ,f ,..., f }…{ 1f

= f 1 f 2 … f n on V. These n-functionals are als

independent over the n-field F = F1 F2 …

{i

i i i

1 2 nf ,f ,...,f } is linearly independent on Vi over the

is true for each i, i = 1, 2, …, n.

Thusin

i i i

j j

j 1

f c f

¦ , i = 1, 2, …, n i.e.

f =1n

1 1

j j

j 1

c f

¦ 2n

2 2

j j

j 1

c f

¦ … nn

n n

j j

j 1

c f

¦ .

in

i i i i k j k k j

k 1f ( ) c f ( )

D D¦

=in

i

k kj

k 1

c

G¦

=i

jc .

This is true for every i = 1, 2, …, n and 1 j n i. I

if each f i is a zero functional i i

jf ( )D = 0 for each j an

scalari

c are all ero Th si i i

f f f are



{n

n n n

1 2 n, ,...,D D D }. B* forms the n-basis

* * *

1 2 nV V ... V .

We prove the following interesting theorem.

THEOREM 1.2.23: Let V = V 1 … V n be a f

nn ) dimensional n-vector space over the n-field

… F n and let B = { 1

1 1 11 2, ,..., nD D D } { 2 2

1 2, ,D D

{ 1 2, ,...,n

n n n

nD D D } be a n-basis for V. Then there

dual basis

B* = { 1

1 1 1

1 2, ,..., n f f f } { 2

2 2 2

1 2, ,..., n f f f } … {

for V* = 1 2 ... nV V V

such that ( )

k

i j f D linear functional f = f 1 f 2 … f n we have

f =1

( )in

i i i

k k

k

f f D

¦ ;

i.e.,

f =1

1 1 1

1

( )n

k k

k

f f D ¦

2

2 2 2

1

( )n

k k

k

f f D ¦

… nn

k ¦

and for each n-vector = 1 2 … n in V

i =

1

( )in

i i i

k k

k

f D D

¦

i.e.,

=1

1 1 1

1

( )n

k k

k

f D D

¦ 2

2 2 2

1

( )n

k k

k

f D D

¦ … 1

nn

k ¦



=1n

1 1

i i

i 1

x

D¦ 2n

2 2

i i

i 1

x

D¦ … nn

n n

i i

i 1

x

D¦

is a n-vector in V then

f j() =1n

1 1 1

i j i

i 1

x f ( )

D¦ 2n

2 2 2

i j i

i 1

x f ( )

D¦ … nn

n

i

i 1

x

¦

=1n

1 1

i ij

i 1

x

G¦ 2n

2 2

i ij

i 1

x

G¦ … nn

n n

i ij

i 1

x

G¦

= 1 2 n

j j jx x ... x ;

so that the unique expression for as a n-linear comk

jD , k = 1, 2, …, n; 1 j n j i.e.

=1n

1 1 1i i

i 1

f ( )

D D¦ 2n

2 2 2i i

i 1

f ( )

D D¦ … nn

ni

i 1

f (

D¦

Suppose Nf = 1

1

f N 2

2

f N … n

n

f N denote the n-nu

the n-dim Nf = dim n

n

f N … dim n

n

f N but dim i

f N

1 = ni –1 so n-dim Nf = (dim V1 –1) dim V2 –1 …1 = n1 – 1 n2 – 1 … nn – 1. In a vectodimension n, a subspace of dimension n – 1

hyperspace like wise in a n-vector space V = V1 (n1, n2, … , nn) dimension over the n-field F = F1

Fn then the n-subspace has dimension (n1 – 1, n2 –

1) we call that n-subspace to be a n-hyperspace of V

is a n-hyper subspace of V.

Now this notion cannot be defined in case of n vect



1, 2, …, n); the n-annihilator of S is S o = 1

oS

linear functionals on V such that f () = 0 0

= f 1 f 2 … f n for every S i.e., = 1

S 1 S 2 … S n. i.e., f i( D i ) = 0 for every i

n. It is interesting and important to note that S

is an n-subspace of V * =* * *

1 2 ... nV V V wh

subspace of V or only just a n-subset of V. If S

0) then S o = V *. If S = V then S o is just the zof V *.

Now we prove an interesting result in case of fin

n-vector space of type II.

THEOREM 1.2.24: Let V = V 1 V 2 … V

space of (n1 , n2 , …, nn ) dimension over the n-fie

… F n of type II. Let W = W 1 W 2 … subspace of V. Then dim W + dim W° = dim V (

(k 1 , k 2 , …, k n )) that is (k 1 , k 2 , …, k n ) + (n1 – k 1 ,

k n ) = (n1 , n2 , …, nn ).

Proof: Let (k 1, k 2, …, k n) be the n-dimension of

… Wn.. Let

P = {1

1 1 1

1 2 k , ,...,D D D } {2

2 2 2

1 2 k , ,...,D D D } …

= P1 P2 … Pn be a n-basis of W. Choose n

{1 1

1 1k 1 n,...,D D } {

2 2

2 2k 1 n,...,D D } … { D

in V such that

B = {1

1 1 1

1 2 n, ,...,D D D } {2

2 2 2

1 2 n, ,...,D D D } … {



+ 1; r = 1, 2, …, n, 1 i k r , because r r

i j ijf ( )D G and

k r + 1 and j k r + 1, from this it follows that for i = 0 whenever Dr is a linear combination r r

1 2a ,a ,

functionalsr r r

r r r

k 1 k 2 na ,a ,...,a are independent for ev

…, n. Thus we must show that they span o

r W , for r =

Supposer *

r f V . Now

r nr r r r

i i

i 1f f ( )f

D¦ so that if f r

i

have r r

if ( ) 0D for i k r and f r =r

r

nr r r

i i

i k 1

f ( )f

D¦ .

We have to show if dim Wr = k r and dim Vr = nr then

nr – k r ; this is true for every r. Hence the theorem.

COROLLARY 1.2.4: If W = W 1 W 2 … W n is k n ) dimensional n-subspace of a (n1 , n2 , …, nn ) dim

vector space V = V 1 V 2 … V n over the n-fiel

F 2 … F n then W is the intersection of (n1 – k 1 )

… (nn – k n ) tuple n-hyper subspaces of V.

Proof: From the notations given in the above theor

W1 W2 … Wn, each Wr is the set of vector f i(

r ) = 0, i = k r + 1, …, nr . In case k r = nr – 1 we s

null space of r nf . This is true for every r, r = 1, 2, …

the claim.

COROLLARY 1.2.5: If W 1 =1 2

1 1 1... nW W W



then each j j

1 2W W for j = 1, 2, …, n, so that W

every j = 1, 2, …,n. Thuso

1W =o

2W . If on the oW2 i.e. 1 2 n

1 1 1W W ... W 1 2

2 2W W ... the two n-subspaces contains a n-vector whic

other. Suppose there is a n-vector 2 = 1 2 is in W2 and not in W1 i.e. 2 W2 and 2 W

corollary just proved and the theorem therefunctional f = f 1 f 2 … f n such that f()

W, but f(2) 0. Then f is in o

1W but not in o

2W

Hence the claim.

Next we show a systematic method of finding t

n-subspace spanned by a given finite n-set 1 2 nn n n

1 1 nF F ... F . Consider a n-system of

linear n-equations, which will be from the poin

n-linear functionals. Suppose we have n-syst

equations.

1 1

1 1 1 1

1 1 1 1

11 1 1n n

1 1 1 1

m 1 1 m n n

A x A x 0

A x A x 0,

!

# #

!

22

2 2 2 2

2 2 2 2

11 1 1n n

2 2 2 2

m 1 1 m n n

A x A x 0

A x A x 0,

!

# #

!

k k kf ( ) k k k kA A thi i t f



k

k k k

i 1 nf (x ,..., x ) =k k

k k k k

i1 1 in nA x ... A x , this is true for

1, 2, …, n; then we are seeking the n-su1 2 nn n n

1 2 nF F ... F ; that is all = 1 2 … k k

if ( )D = 0, i = 1, 2, …, mk and k = 1, 2, …, n. In

we are seeking the n-subspace annihilated by 1

1{f ,

2

2 2 2

1 2 m{f ,f ,..., f } … n

n n n

1 2 m{f ,f ,...,f }. Row reduction

the coefficient matrix of the n-matrix provides systematic method of finding this n-subspace. The

nn) tuple (1

1 1

i1 inA ,...,A ) (2

2 2

i1 inA ,...,A ) … (

gives the coordinates of the n-linear functional k

if , k

n relative to the n-basis which is n-dual to the stand

of 1 2 nn n n1 2 nF F ... F . The n-row space of the n-co

matrix may thus be regarded as the n-space

functionals spanned by (1

1 1 1

1 2 mf , f ,..., f ) ( 2 2 2

1 2 mf ,f ,..., f

(n

n n n

1 2 mf ,f ,...,f ). The solution n-space is the n-s

annihilated by this space of n-functionals.

Now one may find the n-system of equations from

point of view. This is suppose that we are given, mi

1 2 nn n n

1 1 nF F ... F ;

= 1 n

i i...D D = (1

1 1 1

i1 i2 inA A ...A ) … ( n n

i1 i2A A ...A

find the n-annihilator of the n-subspace spanne

vectors. A typical n-linear functional on 1n n

1 1F F

has the form f 1(1

1 1

1 nx ,..., x ) f 2(2

2 2

1 nx ,..., x ) … f n



As in case of usual vector space we in case

spaces of type II define the double dual. Himportant to mention in case of n-vector space

cannot define dual or double dual. This is yet an

between the n-vector spaces of type I and type II

DEFINITION 1.2.14: Let V = V 1 V 2 … V

space over the n-field F = F 1 F 2 … F n (vector space over F i ). Let V* =

* *

1 2 ...V V vector space which is the n-dual of V over the sa

F 1 F 2 … F n. The n-dual of the n-dual spin terms of n-basis and n-dual basis is given in th

Let =

1

2

…

n

be a n-vector in Va n-linear functional L = 1 2

1 2 ... n

n L L LD D D

= 1

1 1( ) ... ( )n

n n L f L f D D

(where f = f 1 f

2

L(f) = 1

1 1( ) ... ( )n

n n L f L f D D

= f() = f 1(1 )

f n(n ), f V * =* *

1 ... nV V f i

*

iV for i =

fact that each i

i LD

is linear is just a reform

definition of the linear operators in iV for each

LD (cf + g) = 1

1 1 1 1( ) ... (n

n n n n L c f g L c f gD D

= ( c1 f 1 + g 1 )(1 ) … (cn f n + g n )(n )

= (c1

f 1

(1

) + g 1

(1

)) … (cn

f n

(n

) + g n

(n

))= (c1 1

1 LD

(f 1 ) + 1

1 LD

(g 1 )) … (cn n

n LD

(f n ) +

= cL( f ) + L(g).

{ n n nD D D } for V = V V V suc



{n1 2 n, ,...,D D D } for V = V1 V2 ... Vn suc

1 2 n

1 1 1, ,...,D D D and let f be a n-linear functional whiceach n-vector in V its first coordinate in the n-ordere

We prove the following interesting theorem.

THEOREM 1.2.25: Let V = V 1 V 2 ... V n be a f

…, nn ) dimensional n-vector space over the n-field F … F n. For each n-vector = 1 2 … n

LD (f) = 1

1 1( ) ... ( )n

n n L f L f D D

= f ( D ) = f 1( D ) …

in V*. The mapping o L , is then an n-isomorphisV**.

Proof: We showed that for each = 1 2 … n-function L= 1 n

1 nL ... LD D

is n-linear. Suppose

… n and = 1 2 … n are in V1 V2 c = c1 c2 … cn is in F = F1 F2 … Fn an

+ i.e., 1 2 … n=(c1

1+

1) (c2

2

+ 2)

+ n). Thus for each f in V*

LJ(f) = f()

= f(c + )

= f 1(c11 + 1) … f n(cnn + n)

= [c1f 1(1) + f 1(1)] … [cnf n(n) + f n(

= 1 1 n

1 1 1 1 1 n n n n

c L (f ) L (f ) ... c L (f ) LD E D E = cL(f) + L(f)

L = cL + L.

F2 Fn If L is a n-linear functional on th



F 2 ... F n. If L is a n linear functional on th

of V then there is a unique n-vector = 1 2

such that L(f) = f () for every f in V*.

The proof is left for the reader to prove.

COROLLARY 1.2.7: If V = V 1 V 2 ... V n is


F n. Each n-basis for V* is the dual of some n-bas

Proof: Given V = V1 V2 ... Vn is a (n1, n

space over the n-field F = F1 F2 ..* *

1 nV ... V is the dual n-space of V over the n-

Let B*= { 11 11 nf ,..., f } . { n

n n1 nf ,...,f } be a n-

an earlier theorem there is a n-basis 1{L

2

2 2 2

1 2 n{L ,L ,...,L }…n

n n n

1 2 n{L ,L ,...,L } for V** =

such that k k

i j ijL (f ) G for k = 1, 2, …, n, 1 i

above corollary for each i, there is a vector k iDk k

iL (f ) = k k

if ( )D for every f k in k V , such th

follows immediately that {1

1 1 1

1 2 n, ,...,D D D } { D

… {n

n n n

1 2 n, ,...,D D D } is a n-basis for V and B*

this n-basis.

In view of this we can say (Wo) o = W.

W = Woo. We can prove this yet in another way. We



p y y

W + dim Wo = dim V. dim Wo + dim (Wo)o = dim V

V = dim V* we have dim W + dim Wo

= dim Wo

which implies dim W = dim Woo. Since W is a subs

we see that W = Woo.

Let V be an n-vector space over the n-field of

define an n-hyper subspace or n-hyperspace of V. W

is (n1, n2, …, nn) dimension over F = F1 F2 …

a n-hyperspace of V i.e. N is of (n1 – 1, n2 – 1, dimension over F then we can define N to be a n-hy

V if

(1) N is a proper n-subspace of V

(2) If W is a n-subspace of V which contains N

W = N or W = V.

Conditions (1) and (2) together say that N is asubspace and there is no larger proper n-subspace in

maximal proper n-subspace of V.

Now we define n-hyperspace of a n-vector space.

DEFINITION 1.2.15: If V = V 1 V 2 … V n is space over the n-field F = F 1 F 2 … F n a n-hyV is a maximal proper n-subspace of V.

We prove the following theorem on n-hyperspace of

THEOREM 1.2.27: If f = f 1 … f n is a non-ze functional on the n-vector space V = V 1 V 2 ...

II over the n-field F = F 1 F 2 … F n , then the n-in V is the n null space of a (not unique) non ze

vector + c where J = 1 … n and c = c1



J

in Nf , c in F = F1 F2

… Fn. Let =

1

in V. Define1 1 n n

1 1 n n

f ( ) f ( ) f ( )c ...

f ( ) f ( ) f ( )

E E E

D D Di.e., each

ci =i i

i i

f ( )

f ( )

E

D, i = 1, 2, … , n;

which makes sense because each f i(

i)0, i = 1,

0. Then the n-vector = – c is in Nf since f(f() – cf() = 0. So is in the n-subspace spanne

Now let N be the n-hyperspace in V. For some

2 … n which is not in N. Since N is a

n-subspace, the n-subspace spanned by N and

space V. Therefore each n-vector in V has th

cD, J = J … Jn in N. c = c1 … cn in F

and N = N1 N2 … Nn where each Ni is a subspace of Vi for i = 1, 2, …, n. The n-vect

scalars c are uniquely determined by . If we ha

c1, 1

in N, c1 in F then (c1 – c) = – 1

if c

would be in N, hence c = c1 and JJ i.e., if iunique n-scalar c such that – c is in N. Call th

It is easy to see g is an n-linear functional on V

n-null space of g.

Now we state a lemma and the proof is left for th

LEMMA 1.2.2: If g and h are n-linear functionah l h

THEOREM 1.2.28: Let V = V 1 V 2 … V n be



space over the n-field F = F 1 F 2 … F n. If { g

2

2 2 2

1{ , , , }r g f f ! … 1{ , , , }n

n n n

r g f f ! be n-linear

on the space V with respective n-null spaces 1{ , N N

2

2 2 2

1{ , ,..., }r N N N … 1{ , ,..., }n

n n n

r N N N . Then (g 1 g

is a linear combination of 1

1 1

1{ ,..., }r f f … 1{n f

if and only if N 1 N

2 … N n contains the n-

{ 1

1 1

1 ... r N N } { 2

2 2

1 ... r N N } … { 1 ...n N

Proof: We shall prove the result for Vi of V1 V

and since Vi is arbitrary, the result we prove is true f

= 1, 2, ., n. Let gi =i i

i i i i1 1 r r c f ... c f and i i

jf ( ) 0D

the clearly gi() = 0. Therefore N

icontains i i

1 2 N N

We shall prove the converse by induction on the

The proceeding lemma handles the case r i = 1. S

know the result for r i = k i-1 and leti

i i i

1 2 k f , f , ,f ! b

functionals with null spacesi

i i

1 k N , , N! such that N

is contained in Ni, the null space of gi. Let (gi)c ( i

1f )

be the restrictions of gi, i

1f , … i

i

k 1f to the subspace

(gi)

c, ( i

1f )c …

i

i

k 1(f ) c are linear functionals on the v

i

i

k N . Further more if i is a vector ini

i

k N and ( i

j(f (D

1, …, k i – 1 then iis in

i

i i i

1 2 k N N ... N and so

Now the result is true for each i, i = 1, 2,



theorem.

Now we proceed onto define the notion of n-tr

linear transformation T = Tn … Tn of the n-

and W of type II.

Suppose we have two n-vector spaces V = V

Vn and W = W1 W2 … Wn over the n-fiel

… Fn. Let T = T1 … Tn be a n-linear transV into W. Then T induces a n-linear transform

into V* as follows:

Suppose g = g1 ...gn is a n-linear functional o

Wn and let f() = f 1(1) … f n(

n) (wher

… n V) f(D) = g(T), that is f() =

gn(Tnn) ; for each i Vi; i = 1,2, … n. The

defines a n-function f from V into F i.e. V1 V2

F1 … Fn namely the n-composition of T, a

V into W with g a n-function from W into F = F

Fn

. Since both T and g are n-linear f is also n-lin-linear functional on V. This T provides us wit

… Tntwhich associates with each n-linear

W = W1 … Wn

a n-linear functional f = Ttg

f n = T1g1 … Tng

n on V = V1 V2 …t i

iT g is a linear functional on Vi. Tt = T1

t …

a n-linear transformation from W*= *1W ...

* *

1 nV ... V for if g1, g2, are in W i.e., g1 = 1

1g

= 1 ng g in W* and c = c1 cn is a n

... F n. For each n-linear transformation T = T 1



from V into W there is unique n-linear transform

1 ...t t nT T from W* = * *

1 ... nW W into V* = V

such that ( ) ( )t

g T g T D D for every g in W* and in

We call T t = 1 ...t t

nT T as the n-transpose of

transformation T t is often called the n-adjoint of T.

It is interesting to see that the following important th

THEOREM 1.2.20: Let V = V 1 V 2 ... V n and W

... W n be n-vector spaces over the n-field F

... F n and let T = T 1 ... T n be a n-linear tran

from V into W. The n-null space of T t

= 1 ...

t

T annihilator of the n-range of T. If V and Wdimensional then

(I) n-rank (T t ) = n-rank T

(II) The n-range of T t is the annihilator of the n

of T.

Proof: Let g = g1 g2 … gn be in W* = W

then by definition (Ttg) = g(T) for each =

1

V. The statement that g is in the n-null space of Tt

g(T) = 0 i.e., g1T11 … gnTn

n = (0 … 0)

V = V1 V2 … Vn .

Thus the n-null space of Tt is precisely the n-an

n-range of T. Suppose V and W are finite dimension

V ( ) d di W ( ) F

For (II), let N = N1 … Nn be the n-null spact



function in the n-range of Tt is in the n-annih

suppose f = Tt

g i.e. f 1

… f n

= Tt1g

1

… g in W* then if is in N; f() = f

1(1) … f

(Tt1g

1)1 … (Ttng

n)n = g(T) = g1(T11)

g(0) = g1(0) … gn(0) = 0 0 … 0. Now

Tt is an n-subspace of the space Nº and dim Nº =

(n2-dimN2) … (nn-dim N

n) = n-rank T = n

the n-range of Tt must be exactly Nº.

THEOREM 1.2.31: Let V = V 1 V 2 … V nW 2 … W n be two (n1 , n2 , …, nn ) and (

dimensional n-vector spaces over the n-field F

F n. Let B be an ordered n-basis of V and B* of V*. Let C be an ordered n-basis of W with dua

Let T = T 1 T 2 … T n be a n-linear transfointo W, let A be the n-matrix of T relative to B

be a n-matrix of T t relative to B*, C*. Then k

ij B

2, …, n. i.e.1 2 n

ij ij ij A A A ! =1 2

ij ij B B !

Proof: Let B = {1

1 1 1

1 2 n, , ,D D D! } { 2 2

1 2, ,D D !

{n

n n n

1 2 n, , ,D D D! } be a n-basis of V. The dual n-b

(1

1 1 1

1 2 nf , f , ,f ! ) (2

2 2 2

1 2 nf ,f , ,f ! ) … ( n n

1 2f ,f ,

(1

1 1 11 2 m, , ,E E E! ) (

2

2 2 21 2 m, , ,E E E! ) … ( n

1 ,E E

n-basis of W and C* = (1

1 1 1

1 2 mg ,g , ,g! ) ( 2

1g ,g

( n n ng g g ) be a dual n basis of C Now b

t k k

k j i(T g )( )D = k t k

j k ig (T )D



k jg

k m

k k pi p

p 1( A )

E¦

=k k m m

k k k

pi j p pi jp

p 1 p 1

A g ( ) A

E G¦ ¦ = k

jiA .

For any n-linear functional f = f 1

… f n

on Vk mk k k k

i i

i 1

f f ( )f

D¦ ; k = 1, 2, …, n.

If we apply this formula to the functional f k = Tt

k gk j

factt k k k

k j i ji(T g )( ) A ,D we have

k nt k k k

k j ji ii 1

T g A f ¦ fro

follows k k

ij ijB A ; true for k = 1, 2, …, n i.e., 1

ijB B

= 1 2 n

ij ij ijA A ... A . If A = A1 A2 … An is

m2 u n2, , …, mn u nn) n-matrix over the n-field F =

… Fn, the n-transpose of A is the (n1 u m1, n2 u mmn) matrix At defined by

1 t 2 t n t

ij ij ij(A ) (A ) ... (A ) = 1 2

ij ijA A ...

We leave it for the reader to prove the n-row r

equal to the n-column rank of A i.e. for each matrixthe column rank of Ai to be equal to the row rank of

…, n.

a. multiplication is associative () = (A f i 1 2



Ai for i = 1, 2, …, n.

b. multiplication is distributive with resp( + ) = + and ( + ) = +

Ai for i = 1, 2, …, n.

c. for each scalar ci F i , ci() = (ci) =i = 1, 2, …, n.

If there is an element 1n = 1 1 … 1 in A1n = i.e., 1n = (1 1 … 1) (1 … n = , In = (1 … n) 1n = (1 … n)

1) = 1 … n = . We call A an n-linear

identity over the n-field F. If even one of the A i’

identity then we say A is a n-linear algebra

identity. 1n = 1 1 … 1 is called the n-iden

The n-algebra A is n-commutative if =

A i.e. if = 1 … n and = 1 … n,

n) (1 … n) = 11 … nn. = (1

… n) = 11 … nn. If each ii =

…, n then we say = for every D, E Awhich DE = to be an n-commutative n-linear

one Ai in A is non commutative we don’t cal

commutative n-linear algebra.

1. All n-linear algebras over the n-field

spaces over the n-field F.

2. Every n-linear algebra over an n-field F

commutative n-linear algebra over F.

where F = F 1 F 2 … F n is the n-field. We call

l i l th fi ld F F F A



polynomial over the n-field F = F 1 … F n. Any e

F[x] will be of the form p(x) = p1(x) p2(x) where pi(x) is a polynomial in F i[x] i.e., pi(x) is a pothe variable x with coefficients from F i; i = 1, 2, …, n

The n-degree of p(x) is a n-tuple given by (n1 where ni is the degree of the polynomial pi(x); i = 1,

We illustrate the n-polynomial over the n-field by an

Example 1.2.8: Let F = Z2 Z3 Z5 Q Z11 b

F[x] = Z2[x] Z3[x] Z5[x] Q[x] Z11[x] is a 5

vector space over the 5-field F. p(x) = x2 + x + 1 x

4x3 + 2x+1 81x7 + 50x2 – 3x + 1 10x3 + 9x2 + 7x

DEFINITION 1.2.18: Let F[x] = F 1[x] F 2[x] …

a n-polynomial over the n-field F = F 1 F 2 …

a n-linear algebra over the n-field F. Infact F[commutative n-algebra over the n-field F. F[x] t

algebra has the n-identity, 1n = 1 1 … 1. W

polynomial p(x) = p1(x) p2(x) … pn(x) to be a polynomial if each pi(x) is a monic polynomial in x f…, n.

The reader is expected to prove the following theorem

THEOREM 1.2.32: Let F[x] = F 1[x] F 2[x] …

n-linear algebra of n-polynomials over the n-field F

… F n . Then

c. fg is n-monic if both f and g are n-monic



d. fg is a n-scalar n-polynomial if and onl scalar polynomials

e. If f + g 0, n-deg (f + g) max (n-deg f

f. If f, g, h are n-polynomials over the n-fi

… F n such that f = f 1(x) f 2(x) 0 … 0 and fg = fh then g = h, wher

g 2(x) … g n(x) and h(x) = h1(x)

hn(x).

As in case of usual polynomials we see in case o

the following. Let A = A1 A2 … An be a with identity 1n = 1 1 … 1 over the n-fiel

… Fn, where we make the convention for any

… n A; º = º1 º

2 … ºn = 1 1

for each A.

Now to each n-polynomial f(x) F[x] = F1[

Fn[x] over the n-field F = F1 F2 … Fn

… n in A we can associate an element

F() =1 2 nn n n

1 1 2 2 n n

i i i i i i

i 0 i 0 i 0

f f f

D D D¦ ¦ ¦! ;

for k = 1, 2, …, n; 1 i nk .

We can in view of this prove the following theor

THEOREM 1 2 33: Let F = F F F be

f(D) =1 2

1 2

1 2( ) ( ) (nnn n

i i n

i i i f f fD D D ¦ ¦ ¦!



1 2

0 0 0

i i i

i i i ¦ ¦ ¦

and for c = c1 c2 … cn F = F1 F2 … (cf + g) = cf() + g(); (fg)() = f()g().

Proof:

f(x) =

1 nm m

1 i n ii i

i 0 i 0f x f x

¦ ¦!

and

g(x) =1 nn n

1 j n j

j j

j 0 j 0

g x g x

¦ ¦!

fg =

1 1 i j 2 2 i j n n

i j i j i ji, j i, j i, jf g x f g x f g

¦ ¦ ¦!

and hence

(fg) =1 1 i j 2 2 i j n

i j 1 i j 2 i

i, j i , j i, j

f g f g f g D D ¦ ¦ ¦!

( = 1 2 … n A, i Ai ; for i = 1, 2, …

=1 1 2 2m n m n

1 i 1 j 2 i 2 j

i 1 j 1 i 2 j 2

i 0 j 0 i 0 j 0

f g f g

§ · § ·§ · § ·D D D D ¨ ¸ ¨ ¸¨ ¸ ¨ ¸

© ¹ © ¹© ¹ © ¹¦ ¦ ¦ ¦ !

n nm n

n i n ji n j n

i 0 j 0

f g

§ ·§ ·D D¨ ¸¨ ¸© ¹© ¹¦ ¦

= f()g()

f ( ) ( ) f ( ) ( ) f ( ) ( )



=1

1 1

n1 1

i ic P¦ 2

2 2

n2 2

i ic P¦ … n

n n

nn n

i ic P¦



1i 1 2i 1 ni 1

then for each jk we have k=1, 2, …, n; 1 jk nk .

k k k k k

k

k k k k k k

j i i j j

i

f (t ) c P (t ) c ¦

true for each k, k = 1, 2, …, n.

Since the 0-polynomial has the properly 0(ti) = = t1 … tn F1 … Fn it follows from the ab

that the n-polynomials {1

1 1 1

0 1 nP ,P , ,P! } { 2 2

0 1P ,P , !

{n

n n n

0 1 nP ,P , , P! } are n-linearly independent. The p

{1, x, …, 1nx } {1, x, …, 2nx } … {1, x, …,

n-basis of V and hence the dimension of V is {n1 + 1

nn + 1}. So the n-independent set {1

1 1

0 nP , ..., P } {

… {n

n n

0 nP , ,P! } must form an n-basis for V. Th

f in V

f =

1

1 1

1

n1 1 1

i i

i 0f (t )P

¦ …

n

n n

n

nn n n

i i

i 0f (t )P

¦ (I

I is called the Lagranges’ n-interpolation formula. Sk jx in I we obtain

1 n

1 k 1 n

1 1 n

1 n

n n j j 1 1 n

i i i

i 0 i 0x ... x (t ) P ... (t ) P

D D

¦ ¦Thus the n-matrix

It can also be shown directly that the n-matrix

when 1 1 1

0 1 n(t , t , , t )! 2 2 2

0 1 n(t , t ,..., t ) … (t



w e10 1 n( , , , )!

20 1 n( , ,..., ) … (

{n1 + 1, n2 + 1, …, nn + 1} set of n-distinct elem

field F = F1 … Fn i.e., each i i

0 1(t , t , , t!

distinct elements of Fi; true for i = 1, 2, …, n.

Now we proceed on to define the new

polynomial function of an n-polynomial over th

F2 … Fn. If f = f 1 … f n be any n-polyn-field F we shall in the discussion

~ ~ ~ ~

1 2 nf f f ... f the n-polynomial n-funct

F taking each t = t1 … tn in F into f(t) where

2, …, n. We see every polynomial function arise

analogously every n-polynomial function aris

way, however if ~ ~f g and f = f 1 … f n an

… gn then ~ ~ ~ ~ ~ ~

1 1 2 2 n nf g ,f g ,..., f g for an

polynomial f and g. So we assume two n-poly

such that f g. However this situation occurs o

field F = F1 F2 … Fn is such that each f

only finite number of elements in it. Suppose polynomials over the n-field F then the product

the n-function ~ ~f g from F into F given by~g (t) for every t = t1 … tn F1 F2 …

Further (fg)(x) = f(x)g(x) hence ( ~ ~f g )(x) =

each x = x1 … xn F1 F2 … Fn.

Thus we see ~ ~f g = ~

fg and it is also a

function We see that the n polynomial function

(c + d)~ = c~ + d ~ and ()~ = ~ ~ for all ,

scalar c, d in the n-field F. Here c = c1 … cn a



, f 1 n

… d n , = 1 … n and = 1 … n then= (c11 + d 1 1 )

~ (c22 + d 2 2 )~ … (cnn + d n

+ d 1 ~1 ) (c2~

2 + d 2 ~2 ) … (cn~

n + d n ~n )

d i i )~ = ci

~i + d i ~i where i , i V i , the compo

space of the n-vector space and i , d i F i; the comp

of the n-field F, true for i = 1, 2, …, n. Further ()~

… (n n )~ = ~1 ~1 … ~

n ~n. The n-mappingcalled an n-isomorphism of A onto A~. An n-isomoonto A~ is thus an n-vector space n-isomorphism ofwhich has the additional property of preserving prod

We indicate the proof of the important theorem

THEOREM 1.2.34: Let F = F 1 F 2 … F n bcontaining an infinite number of distinct elememapping f f ~ is an n-isomorphism of the n-alg

polynomials over the n-field F onto the n-alg polynomial functions over F.

Proof: By definition the n-mapping is onto, and if f

f n and g = g1 … gn belong to F[x] = F1[x] …n-algebra of n-polynomial functions over the f

(cf+dg)~ = cf ~ + dg~ where c = c1, …, cn and d = d1,

F1 … Fn; i.e.,

(cf + dg)~ = ~ ~

1 1 1 1c f d g ~ ~

2 2 2 2c f d g … c

field F. Since f ~ = 0, f k (ti) = 0 for k = 1, 2, …, n

nk and it is implies f = 0.



Now we proceed on to define the new polynomial n-ideal or we can say as n-po

Throughout this section F[x] = F1[x] … Fn[

n-polynomial over the n-field F = F1 … Fn.

We will first prove a simple lemma.

LEMMA 1.2.3: Suppose f = f 1 … f n and d

are any two non zero n-polynomials over the n-fn-deg d n-deg f, i.e., n-deg d = (n1 , …, nn ) an…, mn ), n-deg d n-deg f if and only if each ni

n. Then there exists an n-polynomial g = g 1

F[x] = F 1[x] … F n[x] such that either f – d(f – dg) < n-deg f.

Proof: Suppose f = f 1 … f n

=1 2

1 1 2

1 1 2

1 2

m 1 m 1m i m1 1 2

m i m

i 0 i 0

(a x a x ) (a x

¦ ¦n

n n

n n

n

m 1m in n

m i

i 0

(a x a x )

¦! ;

1 2 n

1 2 n

m m m(a , a ,..., a ) (0, …, 0) i.e., eachi

i

ma 0 fo

d = d1 … dn =

1

1 1

1 1

1

n 1n i1 1

n i

i 0

b x b x

§ ·¨ ¸¨ ¸© ¹

¦ 2

2 2

2 2

2

n 1n i2 2

n i

i 0

b x b x

§ ¨ ¨ ©

¦nn 1§ ·

n n n

n

m m n

n nn

af x d

b

§ ·§ ·¨ ¸ ¨ ¸¨ ¸¨ ¸



nn b¨ ¸¨ ¸

© ¹© ¹

= 0 … 0,

or deg 1 1 1

1

1

m m n

1 11

n

af x d

b

§ ·§ ·¨ ¸ ¨ ¸¨ ¸¨ ¸© ¹© ¹

deg 2 2 2

2

2

m m n

2 2

n

af x

b

§ § ·¨ ̈ ¸¨ ¸¨ © ¹©

n n n

n

nm m n

n nn

n

adeg f x d

b

§ ·§ ·¨ ¸ ¨ ¸¨ ¸¨ ¸© ¹© ¹

< degf 1 degf 2 …

Thus we make take

g = 1 2 n1 1 2 2

1 2 n

1 2 n

m m mm n m n1 2 n

n n n

a a ax x ... x

b b b

§ · § · § · ¨ ¸ ¨ ¸ ¨ ¸¨ ¸ ¨ ¸ ¨ ¸

© ¹ © ¹ © ¹

Using this lemma we illustrate the usual process of lo

of n-polynomials over a n-field F = F1 F2

…

THEOREM 1.2.35: Let f = f 1 … f n , d = d 1 …

polynomials over the n-field F = F 1 F 2 … F n … d n is different from 0 … 0; then the

polynomials q = q1 q2 … qn and r = r 1 … such that

1. f = dq + r; i.e., f = f 1 f 2 … f n = (d 1q

(d nqn + r n ).

2. Either r = r 1 … r n = (0 … 0) or n



be an element of F. Then f is n-divisible by x – c = (x

(xn – cn ) if and only if f(c) = 0 … 0 i.e., f 1(c1



… f n(cn ) = 0 0 … 0.

Proof: By theorem f = (x–c)q + r where r is

polynomial i.e., r = r 1 … r n F. By a theo

earlier we have f(c) = f 1(c1) … f n(cn) = [0q1(c1)

[0q2(c2) + r 2(c2)] … [0qn(cn) + r n(cn)] = r 1(c1)

Hence r = 0 … 0 if and only if f(c) = f 1(c1) …0 … 0.

We know if F = F1 … Fn is a n-field. An eleme

c2 … cn in F is said to be a n-root or a n-zero o

polynomial f = f 1 … f n over F if f(c) = 0 …

f 1(c1) … f n(cn) = 0 … 0.

COROLLARY 1.2.9: A n-polynomial f = f 1 f 2

degree (n1 , n2 , …, nn ) over a n-field F = F 1 F 2

at most (n1 , n2 , …, nn ) roots in F.

Proof: The result is true for n-polynomial of n-degr

1). We assume it to be true for n-polynomials of n-d

1, n2 – 1, …, nn – 1). If a = a1 a2 … an is a n-r

… f n, f = (x – a)q = (x1 – a1)q1 … (xn – an)q

q1 … qn has degree n – 1. Since f(b) = 0, i.e., f 1

f n(bn) = 0 … 0 if and only if a = b or q(b) = q1

qn(bn) = 0 … 0; it follows by our inductive assu

f must have (n1, n2, …, nn) roots.

We now define the n-derivative of a n-polynomi



= 1 1 1m m 1 m

1 1 1 1 1{c m c (x c ) ... (x c ) } 2 2 2m m 1 m

2 2 2 2 2{c m c (x c ) ... (x c ) } …



n n nm m 1 m

n n n n n{c m c (x c ) ... (x c ) }

and this is the statement of Taylor’s n-formula for th

f = 1 nm mx ... x .

If

f =1 2 n

1 2

1 2 n

1 2 n

n n nm m1 2 n

m m m

m 0 m 0 m 0

a x a x ... a x

¦ ¦ ¦1 2

1 1 2 2

1 2

1 2

n nk m k mk 1 2

f m 1 m 2

m 0 m 0

D (c) a D x (c ) a D x (c )

¦ ¦n

n n

n

n

nk mn

m n

m 0a D x (c )

¦

and1 11

1

k k m

1 1 1

k 0 1

D f (c )(x c )

k

¦

2 22

2

k k m

2 2 2

k 0 2

D f (c )(x c )k

¦ … nn

n

km

n n

k 0 n

D f (c )(xk ¦

=1 1 1

1

1 1

k m k 1 1 1m

k m 1

D x (c )(x c )a

k

¦¦

2 2 2

2

2 2

k m k 2 2 2m

k m 2

D x (c )(x c )a

k

¦¦ …

n n nk m kD ( )( )

n n n

n

n n

k m k n n nm

m k n

D x (c )(x c )a

k

¦ ¦



= f 1 … f n.

If c = c1 … cn is a n-root of the n-polynomi

f n the n-multiplicity of c = c1 … cn as a n-ro

f n is the largest n-positive integer (r 1, r 2, …1 nr r

1 n(x c ) ... (x c ) n-divides f = f 1 f 2

THEOREM 1.2.37: Let F = F 1 … F n be a n0) characteristic (i.e., each F i is of characteristi

…, n and f = f 1 … f n be a n-polynomial ov

with n-deg f (n1 , n2 , …, nn ). Then the n-scalar c

is a n-root of f of multiplicity (r 1 , r 2 , …, r n )( 1k D f 1 )(c1 ) ( 2k D f 2 )(c2 ) … ( nk

D f n )(cn ) = 0

0 k i r i – 1; i = 1, 2, …., n. ( )ir

i D f (ci ) 0, for

…, n.

Proof: Suppose that (r 1, r 2, … , r n) is the n-multi c2 … cn as a n-root of f = f 1 … f n. T

polynomial g = g1 … gn such that

f = 1 nr r

1 1 n n(x c ) g ... (x c ) g

and g(c) = g1(c1) … gn(cn) 0 … 0. F

f 1 … f n would be divisible by 1r 1

1(x c )

By Taylor’s n-formula applied to g = g1 …

11 1 m mn r (D g )(c )(x c )ª

=1 1 11 1

1

m r mn r

1 1

m 0 1

D g (x c )

m

¦ …

nn n

n

mn r

n

m 0 n

D g (x c

m

¦



Since there is only one way to write f = f 1 … f

one way to write each component f i of f) as

combination of the n-powers 1k

1 n(x c ) ... (x c ni; i = 1, 2, …, n; it follows that

ik

i i

i

(D f )(c )

k = i i

i i

k r

i ii i

i i

0 if 0 k r

D g (c )r k n.

(k r )!

d d -°®

d d° ¯

This is true for every i, i = 1, 2, …, n. Therefore ik

D0 k i r i – 1; i = 1, 2, …, n and ir D f i(ci) gi(ci) 0

= 1, 2, …, n. Conversely if these conditions are

follows at once from Taylor’s n-formula that th

polynomial g = g1 … gn such that f = f 1

1r

1x c g

1 …

nr

nx c g

nand g(c) = g

1(c

1)

0 0 … 0.

Now suppose that (r 1, r 2, …, r n) is not the largest

integer tuple such that 1r

1x c 2r

2x c …

divides f 1 … f n; i.e., each ir

ix c divides f i

…, n; then there is a n-polynomial h = h1 … h

= 1r 1

1x c

h1 … nr 1

nx c

hn. But this impli



m and f m so f – dq = r = r 1 r 2 … r n m. B

an n-polynomial in m of minimal n-degree we can



deg r < n-deg d so r = 0 … 0. Thus m = dF[x] … dnFn[x]. If g is any other n-monic polynomi

gF[x] = m = g1F1[x] … gnFn[x] then their exists

polynomial p = p1 p2 … pn and q = q1 q2

such that d = gp and g = dq i.e., d = d1 … dn = g

gn pn and g1 … gn = d1q1 … dnqn. Thus d = d

… dn pnqn = d1 … dn and n-deg d = n-deg d n-deg q.

Hence n-deg p = n-deg q = (0, 0, …, 0) and as d

monic p = q = 1. Thus d = g. Hence the claim.

In the n-ideal m we have f = pq + r where p, f

p1 p2 … pn m and f = f 1 … f n m; f =

= (p1q1 + r 1) … (pnqn + r n) where the n-remain

… r n m is different from 0 … 0 and hasdegree than p.

COROLLARY 1.2.10: If p1 , p2 , …, pn are n-polynomia

field F = F 1 … F n not all of which are zero i.e.

there is a unique n-monic polynomial d in F[x] = F

F n[x] and d = d 1 … d n such that

a. d = d 1 … d n is in the n-ideal generated b

where pi = pi1 pi

2 … pin; i = 1, 2, …, n

b. d = d 1 … d n , n-divides each of the n-po

= 1

i p … i

n p i.e. d j / i

j p for j = 1, 2, …, n



… { 1

n p F n[x] … n

n p F n[x]} is called the great

n-divisor or n-greatest common divisor of p1 ,

i l i j ifi d b h di ll



terminology is justified by the proceeding corollary.n-polynomials p1 = 1

1 p … 1

n p , p2 = 2

1 p … 2

n p ,

… n

n p are n-relatively prime if their n-greate

divisor is (1, 1, … , 1) or equivalently if the n

generate is all of F[x] = F 1[x] … F n[x].

Now we talk about the n-factorization, n-irreducib

polynomial over the n-field F.

DEFINITION 1.2.23: Let F be an n-field i.e. F = F 1

F n. A n-polynomial f = f 1 … f n in F[x] = F 1[x]

is said to be n-reducible over the n-field F = F 1 F 2if there exists n-polynomials g, h F[x], g = g 1 …

= h1 … hn in F[x] of n-degree (1, 1, …, 1) s

gh, i.e., f 1 … f n = g 1h1 … g nhn and if such g

not exists, f = f 1 … f n is said to be n-irreducible

field F = F 1 … F n.

A non n-scalar, n-irreducible n-polynomial ove… F n is called the n-prime polynomial over the n

we some times say it is n-prime in F[x] = F 1[x] …

THEOREM 1.2.39: Let p = p1 p2 … pn , f = f 1

f n and g = g 1 g 2 … g n be n-polynomial over t

= F 1 … F n. Suppose that p is a n-prime n-polythat p n-divides the product fg, then either p n-dividdivides g.

relatively prime if f = f 1 … f n and p = p1 and p1 are relatively prime f 2and p2 are relative

i

) 1 f i 1 2 W h ll th1 1



p ) = 1 for i = 1, 2, …, n. We shall prove thaSince (f, p) = (f 1, p1) … (f n, pn) = 1 …

polynomials f 0 = 1

0f … n

0f ; p0 = 1

0 p …

… 1 = f of + po p i.e. 1 … 1 = ( 1 1

0f f + p2 2

0 p p ) … ( n n

0f f + n n

0 p p ) n-multiplying by g

we get g = f 0fg + po pg i.e. (g1

… gn

) = (1

0f f… (

n n n

0f f g +n n n

0 p p g ). Since p n-divides fg it

and certainly p n-divides pp0g. Thus p n-divid

claim.

COROLLARY 1.2.11: If p = p1 … pn is a

divides a n-product f 1 , …, f n i.e. 1 1

1 n f f ! 1 f

1

n n

n f ! . Then p n-divides one of the n-polynom

… 1

n n

n f f ! .

The proof is left for the reader.

THEOREM 1.2.40: If F = F 1 F 2 … F n is

n-scalar monic n-polynomial in F[x] = F 1[x]

be n-factored as a n-product of n-monic primesand only one way except for the order.

Proof: Given F = F1 … Fn is a n-field. F[x

Fn[x] is a n-polynomial over F[x]. Suppose f

is a non scalar monic n-polynomial over the n

…, nn). Thus f and g can be n-factored as n-products

primes in F[x] = F1[x] … Fn[x] and hence f = f

1

1 1

1 m(p p ) n

n n

1 m(p p )

1 1

1 n(q q



= 11 m(p , , p )! … n1 m(p , , p )! = 1 n(q , , q!

n

n n n

1 2 n(q ,q ,...,q ) where1

1 1

1 m(p , , p )! … n n

1 2(p ,p ,

1

1 1

1 n(q ,...,q ) … n

n n n

1 2 n(q ,q ,...,q ) are monic n-prim

F1[x] … Fn[x]. Then1 n

1 n

m m(p ... p ) must n-d

( 1

i

iq … n

n

iq ). Since both 1 n

1 n

m m(p ... p ) and ( 1

i

iqare monic n-primes this means that

t t

t t

i mq p for ev

…, n. Thus we see mi = ni = 1 for each i = 1, 2, …, n

= 1 or ni = 1 for i = 1, 2, …, n. For

n-deg f =1 1 n

1 1 n

1 1 n n

m n m n

1 1 ni j i

i 1 j 1 i 1 j

deg p deg q , , deg p

§ ¨ ¨ © ¦ ¦ ¦ ¦!

In this case we have nothing more to prove. So we m

mi > 1, i = 1, 2, …, n and n j > 1, j = 1, 2, …, n.

By rearranging q

i

’s we can assume i

i i

m n p q

1 1 n

1 1 1 n

1 m 1 m m 1 p , , p ,p , , p ! ! = ( 1 1

1 2q q , …,1 1

1 1

n 1 mq p ,

n n

n n

m mq p ). Thus 1 1 n

1 1 1 n

1 m 1 m m 1 p ,...,p ,p ,...,p = (1

1q , …

n

1q ,…,n

n

n 1q ). As the n-polynomial has n-degree less

…, nn) our inductive assumption applies and shsequence ( 1

1q ,…,1

1

n 1q ,…, n

1q ,…,n

n

n 1q ) is at most rea

of n-sequence ( 1

1 p , …,1

1

m 1 p , … , n

1 p , …, 1

m 1 p ).



If T = T 1 … T n is any n-linear operator on t

space V = V1 Vn We call the n characteri d i h T b h i i l



space V = V 1 … V n. We call the n-characterassociated with T to be n-characteristic roots, n-lateeigen values, n-proper values or n-spectral values.

If T is any n-linear operator and C = C 1 …

n-scalar, the set of n-vectors = 1 … n sucC is a n-subspace of V. It is in fact the n-null spa

linear transformation (T – CI) = (T 1 – C 1 I 1 ) … where I j denotes a unit matrix for j = 1, 2, …, n. We c

C n the n-characteristic value of T if this n-s

different from the n-zero space 0 = 0 0 … 0, fails to be a one to one n-linear transformation a

vector space is finite dimensional we see that (T – CI

one to one and the n-det(T – CI) = 0 0 … 0.

We have the following theorem in view of these prop

THEOREM 1.2.43: Let T = T 1 … T n be a n-line

on a finite dimensional n-vector space V = V 1 …

C = C 1 … C n be a scalar. The following are equ

a. C = C 1 … C n is a n-characteristic valu

… T n.

b. The n-operator (T 1 – C 1 I 1 ) … ( T n – C n Iis n-singular or (not n-invertible).

c. det(T – CI ) = 0 0 … 0 i.e., det(T 1 –

det( T n – C n I n ) = 0 … 0.

field, F = F1 F2 … Fn, if and only if n-de

0 … 0 i.e., det(A1 – C1I1) … det(An –

0) we form the n-matrix (xI – A) = (xI( I A )



… 0), we form the n-matrix (xI A) (xI(xIn – An).

Clearly the n-characteristic values of A in F

Fn are just n-scalars C = C1 … Cn in F =

Fn such that the n-scalars C = C1 … Cn in F

Fn such that f(C) = f 1(C1) … f n(Cn) = 0

For this reason f = f 1 f 2 … f n is called the polynomial of A. Clearly f is a n-polynomial of

in x over different fields. It is important to note

… f n is a n-monic n-polynomial whic

exactly (n1, n2, …, nn). The n-monic polynom

polynomial over F = F1 F2 … Fn.

First we illustrate this situation by the following

Example 1.2.8: Let

1

2 1 0 11 0 1 01 1 0 0 0 4

A 0 1 0 00 2 2 1 1 0

1 0 0 00 0 0 1

0

ª ª º « ª º « » « ª º« » « » « « »« » « » « ¬ ¼« » « »¬ ¼ « ¬ ¼ « ¬

= A1 A2 A3 A4; be a 4-matrix of order (3

2, 5 × 5) over the 4-field F = Z2 Z3 Zh t i ti 4 l i l i t d ith A i

x 6 0 0 0 0

0 x 6 0 0 0

x 1 0 3 x 1 0 6

ª « «

ª º « « » «



x 1 0 3 x 1 0 64 x

0 0 0 x 2 0

0 0 0 0 x

ª º « « » « ¬ ¼ « « ¬

is 4-matrix with polynomial entries.

f = f 1 f 2 … f 4 = det(xI A)

= det (xI1 A1) det (xI2 A2) … det(xI

= {(x + 1)2x + (x + 1)} {(x + 1)(x + 2) × (x

– 2 × 2(x + 1)(x + 2)} (x2 – 4) {(x + 6

1)(x + 2)(x + 4)}.

We see det(xI – A) is a 4-polynomial which is

polynomial and degree of f is (3, 4, 2, 5) over the 4-

Z3 Z5 Z7.

Now we first define the notion of similar n-matrice

entries of the n-matrices are from the n-field.

DEFINITION 1.2.25: Let A = A1 … An be a (n1

× nn ) matrix over the n-field F = F 1 F 2 … F ntakes its entries from the field F i , i = 1, 2, …, n. We

matrices A and B of same order are similar if thernon invertible n-matrix P = P 1 … P n of (n1 × nnn ) order such that

B = P 1 AP where P –1 = 1 1 P P P



1 1

2 2

0 0

0 0

n n

n n

C I

C I

ª º« »« »

« »« »

!

!!

# # #



0 0n n

n n

k k C I

« »« »« »¬ ¼

# # #

!

where t

j I is the t t

j jd d u identity matrix.

From this n-matrix we make the following observa

the n-characteristic n-polynomial for T = T 1 T 2

the n-product of n-linear factors

f = f 1 f 2 … f n

= (x – 1

1C )11d

} (x – 1

1

k C ) 1

nk d

(x – 2

1C )21d

} (

} (x – 1

nC ) 1nd

} (x – n

n

k C )nk n

d .

If the n-scalar field F = F 1 F 2 …

algebraically closed i.e., if each F i is algebraically

= 1, 2, …, n; then every n-polynomial over F = F 1

F n can be n-factored; however if F = F 1 F 2 …n-algebraically closed we are citing a special proper

… T n when we say that its n-characteristic poly

such a factorization. The second thing to be noted i

the number of times t

iC is repeated as a root of f t wh

to the dimension of the space in V t of characteri

associated with the characteristic value t iC ; i = 1, 2

1, 2, …, n. This is because the n-nullity of a n-diagonis equal to the number of n-zeros which has on

LEMMA 1.2.4: Suppose that T = C, where T

is the n-linear operator C = C 1 … C

nwhere

Fi; i = 1 2 n and = 1 is a n



F i; i 1, 2, …, n and 1 … n is a n

f 2 … f n is any n-polynomial then f(T)

f 1(T 1 )1 … f n(T n )n = f 1(C 1 )1 … f n(C n )

It can be easily proved by any reader.

LEMMA 1.2.5: Let T = T 1 T 2 … T n be a n-

on the finite (n1 , n2 , …, nn ) dimensional n-vector

… V n over the n-field F = F 1 … F n. If

^ ` ̂ ` ^ 1 2

1 1 2 2

1 1 1 2

n n

k kC C C C C C ! ! ! !

be the distinct n-characteristic values of T and

2

2

iW } n

n

iW be the n-space of n-characte

associated with the n-characteristic values C i =

n

n

iC . If W = { 1

1W + } +1

1

k W } { 2

1W + }

{ 1

n

W + } + n

n

k W } the n-dim W = ((dim1

1W +(dim 2

1W + } + dim2

2

k W ), } , (dim 1

nW + } +

} dim W n. In fact if t

t

i B is the ordered bas

2, …, k t and t =1, 2, …, n then B = { 1

1 B , } , B

2

2

k B } } { 1

n

B , } , n

n

k B } is an n-ordered n-bas

Proof: We will prove the result for a Wt for th

Suppose that (for each i) we have a vector t

iE

assume that t

1E + } +t

t

k E = 0. We shall show that

each i. Let f be any polynomial. Since tt iT E



y p y t iE proceeding lemma tells us that 0 = f t(Tt) thus 0 = f t(T

f t(Tt)t

t

k E = f t(t

1C ) t

1E + } + f t(t

t

k C )t

t

k E . Choose poly

+ } +t

t

k f such that

t t t

i j ij1 i jf C0 i j

- G ®z¯

.

Then t t t

i t ij j

j

0 f T ,0 G E¦ = t

iE .

Now let t

iB be an ordered basis for t

iW and l

sequence t

t t t

1 k B B , , B ! . Then Bt spans the sub

t

t t

1 k W , , W! . Also Bt is a linearly independent s

vectors for the following reason.

Any linear relation between the vectors in Bt w

form t

t t t

1 2 k 0E E E ! , wheret

iE is some linear c

of the vectors in t

iB . From what we just proved t

i 0E for each i = 1, 2, …, k t. Since each t

iB

independent we see that we have only the trivial lin

between the vectors in Bt.

This is true for each t = 1, 2, …, n. Thus we havedimW = ((dim 1

1W +…+ dim1

1

k W ), (dim 2

1W +… +

(dim n

1W + + dim n

kW ))

characteristic values of T and let W i =1

1

iW W

be the n-null space of T C i I = 1 1( )iT C I

The following are equivalent:



(i) T is n-diagonalizable.

(ii) The n-characteristic n-polynomial for T

is f = f 1 … f n

11 2

1 11

11 1 21 1

kdd d

k x C x C x C ! !

1

1

n

n

d n n

k x C x C ! !

and dim 1 2

1 2, , , !n

n

i i i iW d d d , i = 1, 2, …

(iii) ((dim1

1W +…+ dim1

1

k W ), (dim2

1W +…

(dim 1

nW +…+ dimn

n

k W )) = n-dim V = (n

The n-matrix analogous of the above the

formulated as follows. Let A = A1 … An be

n2 , …, nn× nn ) n-matrix with entries from the n-fi

F n and let ^ ` ̂ `1 2

1 1 1 2

1 2, , , , ! ! !

k kC C C Cthe n-distinct n-characteristic values of A in F. F

=1

1

iW + … +n

n

iW be the n-subspace of all n-col

= X 1 … X

n(with entries from the n-field F =

such that (A C i I)X = (A1 1 1

1

i iC I )X 1 … (A

(0 … 0) = ( 1

1i B …

n

ni B ) and let Bi be a

basis for W i =1

1

iW + } +n

n

iW . The n-basis

Now we proceed on to define the new notion of n-

n-polynomials of the n-linear operator T, the n-

polynomial for T and the analogue of the Cayl



theorem for a n-vector space V over n-field of type

linear operator on V.

In order to know more about the n-linear operat

… Tn

on V1 … V

nover the n-field F = F

1

of the most useful things to know is the class of n-pwhich n-annihilate T = T1 … T

n.

To be more precise suppose T = T1 … T

n

operator on V, a n-vector space over the n-field F =

Fn. If p = p1 … pn is a n-polynomial over the n-

… Fn then p(T) = p1(T1) … pn(Tn) is agaioperator on V = V1 … Vn.

If q = q1 … qn is another n-polynomial ov

n-field F = F1 … F

nthen

(p + q)T = (p)T + (q)T

i.e.,

(p1 + q1)T1 … (pn + qn)Tn = [p1(T1) … pn(Tn)] + [q1(T1) … q

and

(pq)T = (p)T(q)Ti.e.,

(p1q1)T1 … (pnqn)Tn = [p1T1q1T1 … pnT

Therefore the collection of n-polynomials P = P

which n-annihilate T = T1 … T

nin the sense

first n, ( 2

in + 1) operators, (i = 1, 2, …, n) mu

dependent i.e., the first ( 2

1n + 1, 2

2n + 1, }, n

operators are n-linearly dependent i.e., we have



} + 21

1

nC

21n

1T = 0, 2

0C I2 + 2

1C T2 + } + 22

2

nC

22n

2T

n

0C In + n

1C Tn + … + 2n

n

nC

2nn

nT = 0;

That is { 1

0C I1 + 1

1C T1 + … + 21

1

nC

21n

1T } {

} + 22

2

nC

22n

2T } … { n

0C In + n

1C Tn + … + C

… 0 for some n-scalars;1

1

iC ,2

2

iC , …,n

n

iC

i p d n p and p = 1, 2, …, n. Thus the n-ideal of

which n-annihilate T contains a non zero n-po

degree ( 21n , 22n , …, 2nn ) or less.We know that every n-polynomial n-ideal c

multiples of some fixed n-monic n-polynomials

generator of the n-ideal. Thus there corresp

operator T = T1 … Tn a n-monic n-polynom

pn.

If f is any other n-polynomial over the n-fiel

Fn then f(T) = 0 0 … 0 i.e., f 1(T1) … … 0 if and only if f = pg where g = g1

polynomial over the n-field F = F1 … Fn i.e

f n = p1g1 … pngn.

Now we define the new notion of n-polynooperator T: V o V.

DEFINITION 1 2 27 L t T T T b

for the n-linear operator T = T 1 T 2 … T ndetermined by the following properties.

1. p is a n-monic n-polynomial over the n-scal

= F 1 … F n.



2. p(T) = p1(T 1 ) … pn(T n ) = 0 … 0.

3. No n-polynomial over the n-field F = F 1

which n-annihilates T = T 1 T 2 … T nn-degree than p = p1 … pn has.

(n1 × n1 , n2 × n2 , …, nn × nn ) is the order of n-ma … An over the n-field F = F 1 … F n where ni × ni matrix with entries from the field F i , i = 1, 2,

The n-minimal n-polynomial for A = A1

defined in an analogous way as the unique n-monic g

the n-ideal of all n-polynomial over the n-field, F =

F n which n-annihilate A. If the n-operator T = T 1 T 2 … T n is rep

some ordered n-basis by the n-matrix A = A1 …

and A have same n-minimal polynomial because f(T)

… f n(T n ) is represented in the n-basis by the n-m

f 1(A1 ) … f n(An ) so f(T) = 0 … 0 if and only

… 0 i.e., f 1(A1 ) … f n(An ) = 0 … 0 if

f 1(T 1 ) … f n(T n ) = 0 … 0. So f(P -1 AP) = f 1

… f n( 1

n P An P n ) =1

1

P f 1(A1 )P 1 … 1

n P f

P -1 f(A)P for every n-polynomial f = f 1 f 2 … f n. Another important feature about the n-minimal p

of n-matrices is that suppose A = A1 … An is a nn × nn ) n-matrix with entries from the n-field F = F 1Suppose K = K 1 … K n is n-field which contain

THEOREM 1.2.45: Let T = T 1 … T n be a n-

on a (n1 , n2 , …, nn ) dimensional n-vector space

V n

[or let A be a (n1

× n1 , …, n

n× n

n )n-matrix i

An where each Ai is a ni × ni matrix with its



field F i of F = F 1 … F n , true for i = 1, 2, } ,

The n-characteristic and n-minimal n-polyn A] have the same n-roots except for n-multiplicit

Proof: Let p = p1 … pn be a n-minimal n-p= T1 … Tn. Let C = C1 … Cn be a n-

field F = F1 … Fn. To prove p(C) = p1(C1) 0 … 0 if and only if C = C1 … characteristic value of T. Suppose p(C) = p1(C1)

= 0 … 0; then p = (x – C1)q1 (x – C2)

Cn)qn where q = q1 … qn is a n-polynomial, n-deg p, the n-minimal n-polynomial p = p1 q(T) = q1(T1) … qn(Tn) z 0 … 0. Cho

= E1 … En such that q(T)E = q1(T1)E1 …

… 0. Let = q(T)E i.e., = 1 … n

q(Tn)En. Then

0 … 0 = p(T)E = p1(T1)E1 … pn(Tn)

= (T – CI)q(T)E

= (T1 – C1I1)q1(T1)E1 … (Tn –

= (T1 – C1I1)D1 … (Tn – CnIn)

and thus C = C1 … Cn is a n-characteristic

… Tn.

S C C C i h t i

the n-distinct n-characteristic values of T. Then the

n-polynomial for T is the n-polynomial p = p1 …1

1

C ) … (x – 1

1

k

C ) (x – 2

1

C ) … (x – 2

2

k

C ) … (

(x – n

kC ).



(x nk C ).

If = 1 … n is a n-characteristic n-vector

the n-operators {(T1 1

1C I1), …, (T1 1

1

k C I1)}, {(T2

(T2 2

2

k C I2)}, …, {(Tn n

1C In), …, (Tn n

n

k C In)} sen

… n into 0 … 0, thus resulting in {(T1 1

1C I

1

1

k C I1)}, {(T2 2

1C I2), …, (T2 2

2

k C I2)}, {(Tn n

1C In

n

n

k C In)} = 0 … 0 for every n-characteristic n-v

… n.

Hence there exists an n-basis for the underlyiwhich consist of n-characteristic vectors of T = T1

Hence p(T) = p1(T1) … pn(Tn) = {(T1 1

1C I1

1

1

k C I1)} … {(Tn n

1C In), …, (Tn n

n

k C In)} = 0

Thus we can conclude if T is n-diagonalizab

operator then the n-minimal n-polynomial for T is an-distinct n-linear factors.

THEOREM 1.2.46: (CAYLEY-HAMILTON): Let T =T n be a n-linear operator on a finite (n1 , n2 , …, nn ) d

vector space V = V 1 … V n over the n-field F =

F n. If f = f 1 … f n is the n-characteristic, n-polynthen f(T) = f 1(T 1 ) … f n(T n ) = 0 … 0; in oth

n-minimal polynomial divides the n-characteristic

Ti = T11

1

iD … Tnn

n

iD

1 2 n

1 1 1 2 2 2 n

n n n1 1 2 2 n

j i j j i j j

j 1 j 1 j 1

A A A D D ¦ ¦ ¦!



1 2 n j 1 j 1 j 1 ¦ ¦ ¦

ii j1 j nd d , i = 1, …, n. These n-equations may

written in the form

1 2

1 1 1 1 1 1 2 2 2

1 2

n n1 1 2

j i 1 j i i j j i 2 j i

j 1 j 1

T A I T A

G D G ¦ ¦

n

n n n n n n

n

nn n

j i n j i i j

j 1

T A I

G D¦!

= 0 … 0, 1d in d n.

Let B = B1 … Bn denote the element of 1n

1K u

i.e., Bi is an element of i in n

iK u with entriest t

t

i jB

t = 1, 2, }, n. When nt = 2; 1d it, jt d nt.

t t

t t 11 t 21 t

t

12 t t 22 t

T A I A IB

A I T A I

ª º « »

¬ ¼

and det Bt = (Tt – t

11A It)(Tt – t

22A It) – ( t

12A t

21A )

f t is the characteristic polynomial associated with

n. f t = x2-trace Atx + det At. For case nt > 2 it is = ft(Tt) since ft is the determinant of the matrix



Now it tB B = (det Bt)It so that

t

t t t t t t

t

n

t tk i i j k j

i 1

B B det B

G¦

.



t

Therefore

t

t t t

t

nt t

k j j

j 1

0 det B

G D¦ t

t t

k det B D , 1d

Since this is true for each t, t = 1, 2, …, n we0 = (det Bt)

1

1

k D … (det Bn)n

n

k D , 1d k i d ni , i

The Cayley-Hamilton theorem is very imp

useful in narrowing down the search for the

polynomials of various n-operators.

If we know the n-matrix A = A1 …

represents T = T1 … Tn in some ordered n

can compute the n-characteristic polynomial f =

We know the n-minimal polynomial p = p1 …

f i.e., each pi/f i for i = 1, 2, …, n (which we cal

and that the two n-polynomials have the same n-

However we do not have a method of compeven in case of polynomials so more difficult i

roots of the n-polynomials. However if f = f 1 as

f = (x – 1

1C )11d… (x –

1

1

k C )1k 1

d (x – 2

1C )21d

… (x –

(x – n

1C )

n

1

d

… (x – n

n

k C )

nk n

d

{1

1C , …, 1

1

k C } {2

1C

{ n

1C , …,n

n

k C }

Example 1.2.10: Let

A =

1 1 0 0

3 1 11 1 0 0 02 2 1

ª ºª º« » ª « »« » «« »« »



A 2 2 12 2 2 1 1

2 2 01 1 1 0

« « »« » ¬ « »« » ¬ ¼¬ ¼

be a 3-matrix over the 3-field F = Z3 Z5 Q. Clcharacteristic 3-polynomial associated with A is giv

f 2 f 3 = x2 (x – 1)2 (x – 1)(x – 2)2 x2 + 1.

verified that p = p1 … p3 = x2(x – 1)2 (x – 1

(x2 + 1) is the 3-minimal 3-polynomial of A.

Now we proceed on to define the new notion ofsubspaces or equivalently we may call it as i

subspaces.

DEFINITION 1.2.28: Let V = V 1 … V n be a n-v

over the n-field F = F 1 F 2 … F n of type II. L

… T n be a n-linear operator on V. If W = W 1 …n-subspace of V we say W is n-invariant under T if

n-vectors in W, i.e., for the n-vector = 1 …

n-vector T = T 11 … T nn is in W i.e., each T

every i W i under the operator T i for i = 1, 2, T(W) is contained in W i.e., if T i(W i ) is contained in

1, 2, …, n i.e., are thus represented as T 1(W 1 ) …W 1 … W n.

… n then Tw(1 … n) =1

1

wT (1)

Clearly Tw is different from T as domain is W an

V is finite dimensional say (n1, …, nn) diminvariance of W under T has a simple n-matrix

L B B B { 1 1 } {



Let B = B1 … Bn = { 1

1D …1

1

r D } … { D

chosen n-basis for V such that Bc = 1 nB ... Bc c

… { n

1D …n

n

r D } ordered n-basis for W. r =

n dim W.Let A = [T]B i.e., if A = A1 … An then

An = [T1]B1 … [Tn]Bn so that

t

tt t j

1t

nt t t

i j i

i 1

T AD

D¦

for t = 1, 2, …, n i.e.,

TD j = T1

1

1

jD … Tn

n

n

jD

1 2 n

1 1 1 2 2 2 n

1 2 n

n n n1 1 2 2 n

i j i i j i i

i 1 i 1 i 1

A A A

D D ¦ ¦ ¦!

Since W is n-invariant under T, the n-vector TDfor j < r i.e., (j1 < r 1, …, jn < r n).

1 n

1 1 1 n n n

1 n

r r 1 1 n n

j i j i i j i

i 1 i 1

T A A

D D D¦ ¦!

i.e.,k k

k

i jA 0 if jk r k and ik ! r k for every k

Schematically A has the n-block

1 1ª º ªª º

LEMMA 1.2.6: Let W = W 1 … W n be an

subspace of the n-linear operator T = T 1 … T n o

… V n. The n-characteristic, n-polynomial for the n

operator T W =11 ...

nW nW T T divides the n-ch

l i l f T Th i i l l i l f T



polynomial for T. The n-minimal polynomial for T Wn-minimal polynomial for T.

Proof: We have

1 1 n n

1 n

B C B C B CAO D O D O D

ª º ª ª º « » « « »¬ ¼ ¬ ¼ ¬

!

where A = [T]B =1 n1B nBT ... T and > @W B

B Tc

=

Bn =1 1

1wB

Tc

ª º¬ ¼

… n n

nwB

Tc

ª º¬ ¼

. Because of the n-bl

the n-matrix

det(xI A) = det(xI1 A1) det(xI2 A2) … d

where

A = A1 … An

= det(xI B) det(xI D)

= {det (xI1 B1) det(xI1 D1) … det(xIn Dn)}.

That proves the statement about n-ch

polynomials. Notice that we used I = I1 … In to

identity matrix of these n-tuple of different sizes.

The k th power of the n-matrix A has the n-block

Ak = (A1)k … (An)k .

Let T = T1 … Tn be any n-linear opera

…, nn)finite dimensional n-space V = V1 …

W1 … Wn be n-subspace spanned by

characteristic vectors of T = T1 … Tn. Let {

{ 2C 2C } { nC nC } b



{ 2

1C , …,2

2

k C } … { n

1C , …,n

n

k C } be

characteristic values of T. For each i let Wi = W

be the n-space of n-characteristic vectors associ

characteristic value Ci = 11iC … n

niC and let

n

iB } be the ordered basis of Wi i.e., t

iB is a ba

{ 1

1B , …,1

1

k B } … { n

1B , …,n

n

k B } is a n-ord

W = ( 1

1W + … +1

1

k W ) … ( n

1W + … + n

kW

… Wn.In particular n-dimW = {(dim 1

1W + }

(dim 2

1W + … + dim2

2

k W ), …, (dim n

1W + }

We prove the result for one particular Wi = { W

and since Wi is arbitrarily chosen the result is tr

= 1, 2, }, n. Let iBc = { i

1D , …,i

ir D } so that th

form the basis of iBc , the next few 2Bc and so on.

Then Tit

jD = i t

j jt D , j = 1, 2, …, r i where ( i

1t

…, i

1C , …,i

i

k C ,i

i

k C , …,i

i

k C ) where i

jC is re

times, j = 1, }, r i. Now Wi is invariant under TDi in Wi, we have

iD = i i

1 1x D + + i ix D

i

i

1

i

2i

i

r

t 0 0

0 t 0B

0 0 t

ª º« »« »« »« »« »¬ ¼

!

!

# # #!

.



ir « »¬ ¼

The characteristic polynomial of Bi i.e., of iiwT i

i

1

C )i1e … (x –

i

i

k C )

ik 1

ewhere i

je = dim i

jW ; j = 1

Further more gi divides f i, the characteristic polyno

Therefore the multiplicity of i

jC as a root of f i

dim i

jW . Thus Ti is diagonalizable if and only if r i

and only if i

1e + … +i

i

k e = ni. Since what we prov

true for T = T1 … Tn. Hence true for every B1 We now proceed on to define T-n conductor of

W1 … Wn V1 … Vn.

DEFINITION 1.2.29: Let W = W 1 … W n be a n-

subspace for T = T 1 … T n and let = 1 … vector in V = V 1 … V n. The T-n conductor of

the set S T ( D ; W) =1T S ( D 1; W 1 ) …

nT S ( D n;

consists of all n-polynomials g = g 1 … g n over t

= F 1 F 2 … F n such that g(T)D is in W; that is

g 2(T 2 )D 2 … g n(T n )D n W 1 … W n.Since the n-operator T will be fixed throug

discussions we shall usually drop the subscript T and

W) = S(D ;W ) S(D ; W ) The authors usu

LEMMA 1.2.7: If W = W 1 … W n is an n-inv

for T = T 1 … T n , then W is n-invariant

polynomial in T = T 1 … T n. Thus for each

in V = V 1 … V n the n-conductor S( D ; W) =

S( D n; W n ) is an n-ideal in the n-polynomial



( n; n) p y

F 1[x] … F n[x].

Proof: Given W = W1 … Wn V = V1

vector space over the n-field F = F1 … Fn. IEn is in W = W1 … Wn, then TE = T1E1 W = W1 … Wn. Thus T(TE) = T2E = 2

1T E1

in W. By induction Tk E = 1k

1T E1 … nk

nT En i

k. Take linear combinations to see that f(T)E =

f n(Tn)En is in W for every polynomial f = f 1 …The definition of S(D; W) = S(D1;W1) …

meaningful if W = W1 … Wn is any n-subs

a n-subspace then S(D; W) is a n-subspace of F[

Fn[x] because (cf + g)T = cf(T) + g(T) i.e., (c

(cnf n + gn)T1 = c1f 1(T1) + g1(T1) … cnf n(Tn

= W1 … Wn is also n-invariant under T = Tlet g = g1 … gn be a n-polynomial in S(D;

… S(Dn; Wn) i.e., let g(T)D = g1(T1)D1 g

gn(Tn)Dn be in W = W1 … Wn is any n-p

f(T)g(T)D is in W = W1 … Wn i.e.,

f(T)[g(T)D] = f 1(T1) [g1(T1)D1] … f n(Tn

will be in W = W1 … Wn.Since (fg)T = f(T)g(T) we have

(f1g1)T1 (f g )T = f1(T1) g1(T1)

conductor S(D; W) always contains the n-minimal

for T, hence every T-n-conductor n-divides the polynomial for T.

LEMMA 1.2.8: Let V = V 1 … V n be a (n1 ,



dimensional n-vector space over the n-field F = F 1 Let T = T 1 … T n be a n-linear operator on V sn-minimal polynomial for T is a product of n-linear

p1 … pn = (x – 1

1c )

11r

(x – 1

2c )

12r

… (x – 1

1

k c )

1

1k r

(x 2

2c )22r

… (x – 2

2

k c )2

2k r … (x – 1

nc ) 1nr (x – 2

nc ) 2nr

… (

i

i

t c F i , k 1 d t i d k n , i = 1, 2, } , n.

Let W 1 … W n be a proper n-subspace of

which is n-invariant under T. There exist a n-vectorn in V such that

(1) is not in W

(2) (T cI) = (T 1 c1 I 1 )1 … (T n cn I for some n-characteristic value of the n-

Proof: (1) and (2) express that T-n conductor of =

n into W1 … Wn is a n-linear polynomial. Sup

… En is any n-vector in V which is not in W. L

… gn be the T-n conductor of E in W. Then g n-di

… pn the n-minimal polynomial for T. Since E

the n-polynomial g is not constant. Therefore g = g1

= (x – 1

1c )11e … (x –

1

1

k c )1k 1

e (x – 2

1c )21e … (x –

2

2

k c )2k

2e

n

(T1 1

1

jc jI1)D1 … (Tn n

n

jc jIn)

= (T1 1

1

jc I1)h1(T1)E1 … (Tn n

n

jc In)

= g1(T1)E1 g2(T2)E2 … gn(Tn

with gi(Ti)Ei Wi for i = 1, 2, }, n.



Now we obtain the condition for T to be n-tr

THEOREM 1.2.47: Let V = V 1 … V n be a fin


and let T = T 1 … T n be a n-linear operator

n-triangulable if and only if the n-minimal polyn

n-product of n-linear polynomials over F = F 1

Proof: Suppose the n-minimal polynomial p = pfactors as p = (x – 1

1c )11r … (x –

1

1

k c )1k 1

r (x – 2

1c )

… (x – 2

2

k c )2k 2

r … (x – n

1c )n

1r …(x – n

n

k c )nk n

r .

application of the lemma just proved we arrive a

basis. B = {1

1D …1

1

nD } {1

2D …2

2

nD } …

B1 B2 … Bn in which the n-matrix repres

… Tn is n-upper triangular.

1 n1 B n B[T ] ... [T ]

1

1

1 1 1 2 2 211 12 1n 11 12 1

1 1 2 2

22 2n 22 2

a a a a a a

0 a a 0 a a

ª º ª « » « « » «

« » «

! !

Merely [T]B = the n-triangular matrix of (n1 × n1,

order shows that

T j = T11

1

jD … Tnn

n

jD



=1

1 1

1j 1a D +…+1 1 1

1 1

j j ja D 2

2 2

1j 1a D +…+2 2

2

j ja

n

n n

1j 1a D + … +n n n

n n

j j ja D

1d jid ni; i = 1, 2, }, n, that is TD j is in the n-subspa

by { 1

1D …1

1

jD } … { n

1D …n

n

jD }. To find { 1

1D …

{ n

1D …n

n

jD }; we start by applying the lemm

subspace W = W1 … Wn = {0} … {0} to ovector 1

1D … n

1D . Then apply the lemma to W

… n

1W , the n-space spanned by D1 = 1

1D …

obtain D2 =1

2D … n

2D . Next apply lemm1 n

2 2W ... W , the n-space spanned by D

and 1 n2 2...D D . Continue in that way. After D1, D

have found it is the triangular type relation given by

for ji = 1, 2, …, ni, i = 1, 2, …, n which proves

subspace spanned by D1, D2, …, Di is n-invariant undIf T is n-triangulable it is evident that the n-ch

polynomial for T has the form f = f 1 … f n = (x –

– 1

1

k c )1k 1

d… (x– n

1c )n1d

… (x – n

n

k c )nk n

d. The n-diag

1 1 2 2

COROLLARY 1.2.12: If F = F 1 … F n is an closed n-field. Every (n1 × n1 , …, nn × nn ) n-m

similar over the n-field F to be a n-triangular ma

Now we accept with some deviations in the

n algebraically closed field when ever F is the co



n-algebraically closed field when ever Fi is the co

THEOREM 1.2.48: Let V = V 1 … V n be


and let T = T 1 … T n be a n-linear operator

V n. Then T is n-diagonalizable if and only if polynomial for T has the form

p = p1 … pn

=1

1 1

1 1( ) . . . ( ) . . . ( ) . . . ( n

kc x c x c x

where 11 11 1{ , . . ., } . . . { , . . ., } n

n nk k c c c c are n-dist

F = F 1 … F n.

Proof: We know if T = T1 … Tn is n-diag

minimal polynomial is a n-product of n-distinc

Hence one way of the proof is clear.To prove the converse let W = W1 …

subspace spanned by all the n-characteristic n-v

suppose W z V. Then we know by the prope

operator that their exists a n-vector D = D1 …

not in W and the a n-characteristic value1

1

j jc c

such that the n-vector

E = (T – c jI)D1 n(T c I ) (T c I )D D

for every n-polynomial h. Now

p = (x – c j)q

= p1 … pn

=1 n

1 n j 1 j n(x c )q . . . (x c )q

for some n-polynomial q = q1 … qn. Also



q – q(c j) = (x – c j)h,

i.e.,1 1 1

1 1 1

1 1 j j jq q (c ) (x c )h , …,n

n n jq q (c ) (x

We have

q(T)D – q(c j)D = h(T) (T – c jI)D = h(T)EBut h(T)E is in W = W1 … Wn and since

0 = p(T)D = (T – c jI) q(T)D= p1(T1)D1 … pn(Tn)Dn

=1 n

1 n

1 j 1 1 1 1 n j n n(T c I )q (T ) . . . (T c )q (T ) D

and the n-vector q(T)D is in W i.e., q1(T1)D1 … in W. Therefore

1 n

1 n

j 1 j 1 n j nq(c ) q (c ) . . . q (c )D D D

is in W. Since D = D1 … Dn is not in W, we h

1 n

1 n

1 j n jq (c ) . . . q (c ) = 0 … 0.

This contradicts the fact that p has distinct roots.claim.

We can now describe this more in terms of how

are determined and its relation to Cayley Hamilton T

n-vector spaces of type II. Suppose T = T1 … linear operator on a n-vector space of type I

represented by the n-matrix A = A1 … An in son-basis for which we wish to find out whethe

diagonalizable. We compute the n-characteristic pol

(T – c1I) … (T – ck I) =

1

1 1 n

1 1 1 1 k 1 n 1 n(T c I ) . . . (T c I ) . . . (T c I ) . . .

is the n-zero operator.Here we have some problems about s

algebraically closed n-field in a universal sens



speak of the n-algebraically closed n-field r

polynomial over the same n-field. Now we proc

or introduce the notion of simultaneous n-diag

simultaneous n-triangulation. Throughout this se… Vn will denote a (n1, n2, …, nn) finite n-ve

the n-field F = F1 … Fn and let = 1 …

family of n-linear operator on V. We now discus

simultaneously n-triangularize or n-diagonalize

in F. i.e., to find one n-basis B = B1 … Bn

matrices [T]B = 1 n1 B n B[T ] [T ] ! , T in (or n-diagonal). In case of n-diagonalization it is

be in the commuting family of n-operators UT =

… UnTn = T1U1 … TnUn for all T

follows from the simple fact that all n-diago

commute. Of course it is also necessary that eac

be an n-diagonalizable operator. In order to si

triangulate each n-operator in we see each n-o

n-triangulable. It is not necessary that be

family, however that condition is sufficient for

triangulation (if each T = T1 … Tn can be

triangulated).We recall a subspace W = W1 … W

under if W is n-invariant under each operato

b. for each T = T 1 … T n in = 1 …

vector T D = T 1D 1 T 2D 2 … T nD n i

subspace spanned by D = D 1 … D n a

… W n.

Proof: Without loss in generality let us assume that



Proof: Without loss in generality let us assume that

only a finite number of n-operators because of this o

Let {T1, …, Tr} be a maximal n-linearly independe

of ; i.e., a n-basis for the n-subspace spanned by … Dn is a n-vector such that (b) hold

i i i

1 nT T .. . T ,1 i r d d then (b) will hold fo

operator which is a n-linear combination of T1, …, T

By earlier results we can find a n-vector1E E

in V (not in W) and a n-scalar 1 1 1

1 nc c . . . c 1 1 1(T c I) E is in W; i.e., 1 1 1 1

1 1 1 1 n(T c I ) . . . (T E

W1 … Wn i.e., each 1 1 1

r r r r(T c ) W ; E r = 1, 21 1 1

1 nV V .. . V be the collection of all vector Ethat (T

1– c

1I)E is in W. Then V

1is a n-subspace of

properly bigger than W. Further more V1

is n-invarifor this reason. If T = T1 … Tn com

1 1 1

1 nT T .. . T then 1 1 1 1(T c I)T T(T c I) E E .

V1 then (T1 – c1I)E is in W i.e., TE is in V1 for all

for all T in . Now W is a proper n-subspace 2 2 2

1 nU U .. . U be the n-linear operator on V1

restricting2 2 2

1 nT T . . . T to the n-subspace

minimal polynomial for U2 divides the n-minimal

Let V2 be the set of all n-vectors E in V1 such t

is in W. Then V2

is n-invariant under . Applyin

to3 3 3

1 nU U .. . U the restriction of T

3to V

2

in this way we shall reach a n-vector D =r r r

1 n. . .E E E such that j j

(T c I) D is in W; j



The following theorem is left as an exercise for t

THEOREM 1.2.49: Let V = V 1 … V n be a fdimension n-vector space over the n-field Fcommuting family of n-triangulable, n-linear op

V 1 … V n. There exists an n-ordered basis f

V n such that every n-operator in is reptriangular n-matrix in that n-basis.

In view of this theorem the following corollary i

COROLLARY 1.2.13: Let be a commuting famn2 u n2 , …, nn u nn ) n-matrices (square), i.e., (

matrices) over a special algebraically closed n

… F n. There exists a non-singular (n1 u n1 , n

nn ) n-matrix P = P 1 P 2 … P n with entries

F n such that 1 1 1

1 1 1 . . . n P A P P A P P A

triangular for every n-matrix A = A1 … An i

Next we prove the following theorem.

THEOREM 1.2.50: Let = 1 … n b

dimensional space. Choose any T = T1 … Tn in

not a scalar multiple of n-identity.

Let1 2 n

1 1 2 2 n n

1 k 1 k 1 k{c , . . ., c } {c , . . ., c } . . . {c , . . ., c

n-character values of T and for each (i(t)) let t

iW

space of the n-null space i 1 n

i iW W ... W ; t = 1



p p i i

(T – ciIi) = (T1 – C1I1) … (Tn – cnIn). Fix i,

invariant under every operator that commutes wt

i{ } be the family of linear operators ont

iW orestricting the operators in t where = 1 … n to the subspace t

iW . Each operator in t

i is dia

because its minimal polynomial divides the minimal

for the corresponding operator in t since dim t

iW <

operators int

i can be simultaneously diagonalized.In other words

t

iW has a basis t

iB which consist

which are simultaneously characteristic vectors

operator t

i . Since this is true for every t, t = 1, 2, …

T = T1 … Tn is n-diagonalizable and

1 2

1 1 2 2 n

1 k 1 k 1B {B , . . ., B } {B , . . ., B } . . . {B , . . . is a n-basis for the n-vector space V= V1 … Vn.

Let V = V1 … Vn be a n-vector space over the

F1 … Fn. Let1

1 1 n

1 k 1W {W , .. . W } .. . {W , .

n-subspaces of the n-vector space V. We say that

are independent if t

1 t t t

1 k i i. . . 0, in WD D D impli

t

i 0D i.e., if 1 1 n n

1 k 1 k( . . . ) . . . ( . . . )D D D D

LEMMA 1.2.10: Let V = V 1 … V n be a finite d

n2 , …, nn ) n-vector space of type II. Let 1

1{ , .W

1{ , . . ., }n

n n

k W W be n-subspaces of V and let W = W

1 2

1 1 2 2

1 1 1. . . . . . . . . . . . n

k kW W W W W

following are equivalent




a.1

1 1

1 1{ , . . ., } . . . { , . . ., } n

n n

k k W W W W are n-in

1{ , . . ., }t

t t k W W are independent for t = 1, 2, …

b. For each jt , 2 d jt d k t , t = 1, 2,

11( . . . )

t t

t t t

j jW W W = {0} for t = 1, 2, …,

c. If t

i B is an ordered basis for t

iW , 1 d i d k

then the n-sequence 1

1 1

1 1{ , . . ., } . . . { k B B Bn-ordered basis for n-subspace W = W 1

2

1 1

1 1. . . . . . . . . n

n n

k k W W W W .

Proof: Assume (a) Lett

t t t

j 1W (W .. . WD

are vectors t 1t t1 j, . . . , D D with t t

i iWD such that

Dt = 0 + … + 0 = 0 and sincet

t t

1 k W , .. ., W are

must be thatt 1

t t t t

1 2 j. . . 0

D D D D . This is

t = 1, 2, …, n. Now let us observe that (b) impl

0 = t

t t t t

1 k i i. . . ; WD D D ; i = 1, 2, …, k t. (We zero vector and zero scalar by 0). Let jt be the l

such thatt

i 0.D z Then t t t

i j j0 . . . ; 0 D D D z

Any linear relation between the vector in Bt

w

formt

t t

1 k . . .E E = 0 wheret

iE is some linear com

vectors in

t

iB . Since t

t t

1 k W , .. . , W are independent e0. Since each t

iB is an independent relation. T

between the vectors in Btis trivial. This is true for ev



2, …, n; so in

B = B1 … B

n

=1 n

1 1 n n

1 k 1 k {B , . . ., B } . . . {B , . . ., B }

every n-relation in n-vectors in B is the trivial n-relat

It is left for the reader to prove (c) implies (a).

If any of the conditions of the above lemma h

the n-sum W =1

1 1 n

1 k 1(W .. . W ) . . . (W .. .

direct or that W is the n-direct sum of 1

1 k

{W , .. ., W

n

n n n

1 2 k {W , W , . . ., W } and we write

1

1 1 n

1 k 1W (W .. . W ) . . . (W . . . W

This n-direct sum will also be known as the n-i

sum or the n-interior direct sum of 1

1 1

1 k {W , .. . , W

n

n n

1 k {W , .. ., W }.

Let V = V1 … Vn be a n-vector space over t

= F1 … Fn. A n-projection of V is a n-linear ope

… En on V such that E2 = 2 2

1 nE ... E = E1

E.Since E is a n-projection. Let R = R1 … R

range of E and let N N N be the null spac

3. The unique expression for D as a sum o

and N is D = ED + (D – ED), i.e., D1 …

(D1 – E1D1) … EnDn + (Dn – EnDn).

From (1), (2) and (3) it is easy to verify. If R =

and N = N1 … Nn are n-subspaces of V suc



p

N = R1 N1 … Rn Nn there is one a

projection operator E = E1 … En which ha

n-null space N. That operator is called the n-p

along N. Any n-projection E = E1 … En

diagonalizable. If 1

1 1 n

1 r 1{ , . . . , } . . . { , . . .,D D D D

of R and ^ ` ^ `1 1 n n

1 1 n n

r 1 n r 1 n,..., ... ,..., D D D D a n-b

the n-basis B =1 n

1 1 n n

1 n 1 n{ , . . . , } . . . { , . . .,D D D D

Bn, n-diagonalizes E = E1 … En.

1 nB 1 B n B[E] [E ] [E ] !

1 nI 0 I 0. . . ,

0 0 0 0

ª º ª º « » « »

¬ ¼ ¬ ¼

where I1 is a r1 u r1 identity matrix so on andidentity matrix.

Projections can be used to describe the

decomposition of the n-space V = V1 … Vn

V = 1

1 1 n

1 k 1(W .. . W ) . . . {W .. . for each j(t) we define

t

jE on Vt. Let D be in V

sum of vectors from the spaces t

iW with i z j. In t

projections t

jE we have Dt = t t t t

1 k E . . . ED D for e

The above equation implies;t t

t 1 k I E . . . E . Note z j then t t

i jE E = 0 because the range of t

jE is the su

which is contained in the null space of tE This is tr



which is contained in the null space of iE . This is tr

t, t = 1, 2, …, n. Hence true on the n-vector space

V =1

t 1 n

i k 1 k(W . . . W ) . . . (W .. . W

Now as in case of n-vector spaces of type I we i

vector spaces of type II obtain the proof of the

theorem.

THEOREM 1.2.51: If V = V 1 … V n is a n-vect

type II and supposeV =1

1 1

1 1( . . . ) . . . ( . . . n

k kW W W W

then there exists (k 1 , …, k n ); n-linear operators 1

1{ E

… 1{ , . . ., }n

n n

k E E on V such that

a. each t

i E is a projection, i.e., 2( )t

i E = t

i E f

…, n; 1 d i d k t .

b. t t

i j E E = 0 if i z j; 1 d i, j d k t ; t = 1, 2, …,

c. I = I 1 … I n

=1

1 1

1 1( . . . ) . . . ( . . . n n

k k E E E E

d. The range of

t

i E is

t

iW i = 1, 2,…, k t and t =

Proof: We are primarily interested in n-direc

vectors1 n

1 1 n n

1 k 1 k { , . . ., } . . . { , . . ., }D D D D with

that1 n

1 1 n n

1 k 1 k. . . . . . . . .D D D D D and t

1 11 1 1 1 n n1 1 k k 1 1T T .. . T . . . T . . .D D D D

We shall describe this situation by saying that T

sum of the operators1 1 n

1 k 1{T , . . ., T } . . . {T , .



sum of the operators11 k 1{T , . . ., T } . . . {T , .

be remembered in using this terminology that th

linear operators on the space V = V1 … various n-subspaces

V = W1 … Wn.

=1

1 1 n n

1 k 1 k(W .. . W ) . . . (W .. . W

which enables us to associate with each D = D1

a unique n, k-tuple 11 1 n1 k 1( , . . ., ) . . . ( , . . .,D D D

t

iD t

iW ; i = 1, 2, …, k t; t = 1, 2, …, n.

1

t 1 n

i k 1( . . . ) . . . ( . . .D D D D

in such a way that we can carry out the n-linear

by working in the individual n-subspacesi

W The fact that each Wi is n-invariant under T ena

the action of T as the independent action of the

the n-subspaces t

iW ; i = 1, 2, …, k t and t = 1

purpose is to study T by finding n-invarian

decompositions in which the t

iT are operators o

nature.

be as before. Then a necessary and sufficient con

each n-subspace t

iW ; to be a n-invariant under T t fo

is that t t

i t i t E T E T or TE = ET for every 1 d i d k t a

…, n.

Now we proceed onto define the notion of



p

decomposition for n-vector spaces of type II over the

F1 … Fn.

Let T be a n-linear operator of a n-vector space V Vn of (n1, n2, …, nn) dimension over the n-field F

… pn be a n-minimal polynomial for T; i.e.,

p =11 nk1 1 1

1

rr r1 1 n

1 k 1(x c ) . . . (x c )x . . . (x c ) . . . (x

where1 n

1 1 n n

1 k 1 k {c , . . ., c } . . . {c , . . ., c } are distinct el

= F1 … Fn, i.e.,t

t t1 k {c , .. ., c } are distinct elemen

1, 2, …, n, then we shall show that the n-space V =

Vn is the n-direct sum of null spacessirs

s i s(T c I ) , i =

and s = 1, 2, …, n.

The hypothesis about p is equivalent to the fact

triangulable.

Now we proceed on to give the primary n-dec

theorem for a n-linear operator T on the finite dim

vector space V = V1 … Vn over the n-field F =

Fn of type II.

THEOREM 1.2.53: (PRIMARY N-DECOMPOSITION TH

T = T T be a n linear operator on a finite

i = 1, 2, …, n. Then

i. V = W 1 … W n

=1

1 1

1 1( . . . ) . . . ( . n

kW W W

ii. each 1 . . . n

i i iW W W is n-invariant u

…, n.

iii if r T is the operator induced on r W



iii. if iT is the operator induced on iW

minimal polynomial for r

iT is r = 1, 2,

…, n.

The proof is similar to that of n-vector spac

hence left as an exercise for the reader. In view

we have the following corollary the proof of

direct.

COROLLARY 1.2.14: If 1

1 1

1{ , . . ., } . . . { k E E E

the n-projections associated with the n-primary

of T = T 1 … T n , then each t

i E is a polynom

d k t and t = 1, 2, …, n and accordingly if a lincommutes with T then U commutes with each of

subspace W i is invariant under U.

Proof: We can as in case of n-vector spaces of case of n-linear operator T of type II, the notio

part of T and n-nilpotent part of T.

Consider the n-minimal polynomial for T

product of first degree polynomials i.e., the case

pi is of the form t t

i ip x c . Now the range of

D =1 1

1 1 1 1

1 1 k k (c E . . . c E ) . . . n n

1 1(c E . . .

so

N =1 1

1 1 1 1

1 1 1 1 1 k 1 k {(T c I ) E . . . (T c I ) E }

1 n

n n n n

n 1 n 1 n k n k{(T c I ) E . . . (T c I ) E }



The reader should be familiar enough with n-projecti

so that

N2 =1 1

1 2 1 1 2 1

1 1 1 1 1 k 1 k{(T c I ) E . . . (T c I ) E }

n

n 2 n n 2 n

n 1 n 1 n k n k{(T c I ) E . . . (T c I ) E

and in general,

Nr

= 1 1

1 1

r r1 1 1 1

1 1 1 1 1 k 1 k{(T c I ) E . . . (T c I ) E } n n

n

r rn n n

n 1 n 1 n k n{(T c I ) E . . . (T c I ) E

When r t ri for each i we have Nr

= 0 because

n-operators trt

t i t(T c I ) = 0 on the range of t

iE ; 1 d= 1, 2, …, n. Thus (T – c I)r = 0 for a suitable r.

Let N be a n-linear operator on the n-vector spac

… Vn. We say that N is n-nilpotent if there is sominteger (r1, …, rn) such that ir

iN = 0. We can choose

1, 2, …, n then Nr = 0, where N = N1 … Nn.

In view of this we have the following theorem fspace type II, which is similar to the proof of n-vect

type I.

THEOREM 1.2.54: Let T = T 1 … T n be a n-line

th ( ) fi it di i l t

The n-diagonalizable operator D and operator N are uniquely determined by (i) and them is n-polynomial in T 1 , …, T n.

Consequent of the above theorem the follow

direct.



COROLLARY 1.2.15: Let V be a finite dimen

space over the special algebraically closed field

F n. Then every n-linear operator T = T 1 … … V n can be written as the sum of a n

operator D = Di … Dn and a n-nilpotent o

… N n which commute. These n-operatorunique and each is a n-polynomial in (T 1 , …, T n )

Let V = V1 … Vn be a finite dimensional

over the n-field F = F1 … Fn and T = T1 arbitrary and fixed n-linear operator on V. If Dis n-vector in V, there is a smallest n-subspace o

invariant under T and contains D. This n-su

defined as the n-intersection of all T-invaria

which contain D; however it is more profitable

for us to look at things this way.

If W = W1 … Wn be any n-subspace

space V = V1 … Vn which is n-invaria

contains D = D1 … Dn i.e., each Ti in T

subspace Wi in Vi is invariant under Ti and contai = 1, 2, …, n.

Then W must also contain TD i.e., TiDi is in



(iii) If U = U 1 … U n is a n-linear operatorinduced by T, then the n-minimal polynom

pD .

Proof: Let g = g1 … gn be a n-polynomial ove

F = F1 … Fn. Write g = pDq + r, i.e., g1 …

h f



+ r1 …nn np qD + rn where pD =

11p D … nnp D f

… Dn, q = q1 … qn and r = r1 … rn so gi

true for i = 1, 2, …, n. Here either r = 0 … 0 or

deg pD = (k 1, …, k n).

The n-polynomial pDq =11p D q1 …

nnp D qn

n-annihilator of D = D1 … Dn and so g(T)D =

g1(T1)D1 … gn(Tn)Dn = r1(T1)D1 … rn(Tn)D

r1 … rn = 0 0 … 0 or n-deg r < (k 1, k 2, …vector r(T)D = r1T1(D1) … rnTn(Dn) is

combination of the n-vectors D, TD, …, Tk–1D; i.e

combination on n-vectors D = D1 … Dn.

1 1 n nT T .. . T ,D D D2 2 2

1 1 n nT T . . . TD D D , …,1 nk 1 k 1k 1

1 1 n nT T .. . T D D D

and since g(T)D = g1(T1)D1 … gn(Tn)Dn is a

vector in Z(D; T) i.e., each gi(Ti)Di is a typical vector

for i = 1, 2, …, n. This shows that these (k 1, …, k nspan Z(D; T).

These n-vectors are certainly n-linearly in

because any non-trivial n-linear relation between t

i l i l h h (T)( )

= g(T) pD(T)D=

1 n1 1 1 1 1 n n ng (T ) p (T ) . . . g (T ) pD DD

= g1(T1)0 … gn(Tn)0

= 0 … 0.

Thus the n-operator pDU = p1D1(U1) … t i Z( T) Z( T ) Z(



every n-vector in Z(D; T) = Z(D1; T1) … Z(

0 … 0 and is the n-zero operator on Z(D; T

if h = h1 … hn is a n-polynomial of n-degr…, k n) we cannot have h(U) = h1(U1) … hn(

0 for then h(U)D = h1(U1)D1 … hn(Un)D n) =

hn(Tn)Dn = 0 … 0; contradicting the d

This show that pD is the n-minimal polynomial fo

A particular consequence of this nice theore

D1 … Dn happens to be a n-cyclic vector forTn then the n-minimal polynomial for T must

equal to the n-dimension of the n-space V =

hence by the Cayley Hamilton theorem for n-ve

have that the n-minimal polynomial for T is the

polynomial for T. We shall prove later that for

n-vector D = D1 … Dn in V = V1 … Vn-minimal polynomial for T = T1 … annihilator. It will then follow that T has a n-cyc

only if the n-minimal and n-characteristic polyn

identical. We now study the general n-operator

Tn by using n-operator vector. Let us cons

operator U = U1 … Un on the n-space W =

of n-dimension (k 1, …, k n) which has a cyclic n-

If we let i i 1U D D ; i.e., i i i

1 n. . .D D D and

implies 1 ni 1 i 1i i

1 1 1 n n nU , . . ., U

D D D D ; 1 d i d k i –

action of U on the ordered n-basis 1

1 1

1 k { , . . ., }D D n

n n

1 k { , . . ., }D D is UDi = Di+1 for i = 1, 2, …,

ti 1i

t t tUD D for i = 1, 2, …, k t – 1 and t = 1, 2, …,

1 k



c0D1 – … –ck–1D

k i.e.,

t

k t 1 t k

t t 0 t k 1 tU c .. . c D D D

for t = 1, 2, …, n, where

pD = 1 1

1

k 1 k 1 1 1

0 1 k 1{c c x . . . c x x }

… n n

n

k 1 k n n n

0 1 k 1{c c x . . . c x x }

.

The n-expression for UDk follows from the fact

… 0 i.e.,

1 n1 1 1 1 n np (U ) . . . p (U )D DD D = 0 … 0.

i.e.,k k 1

k 1 1 0U c U .. . c U c DD D D = 0 …

i.e.,1 1

1

k k 11 1 1

1 1 k 1 1 1 1 1 1 0 1

U c U .. . c U c

D D D D 2 2

2

k k 12 2 2

2 2 k 1 2 2 1 2 2 0 2U c U .. . c U cD D D D

n n

n

k k 1n n n

n n k 1 n n 1 n n 0U c U .. . c U c

D D D D

= 0 … 0.

This is given by the n-matrix of U = U1 … Uordered basis

B = B1 B

n

n

0

n

1

n

2

n

k 1

0 0 0 0 c

1 0 0 0 c

0 1 0 0 c

0 0 0 1 c

ª º« »

« »« »

« »« »« »¬ ¼

!

!

!

# # # # #

!

.



The n-matrix is called the n-companion n-matrn-polynomial1 n1 np p . . . pD D D or can also

with some flaw in notation as1 n

1 np . . . pD D wh

pn.

Now we prove yet another interesting theore

THEOREM 3.2.56: If U = U 1 … U n is a n-

on a finite (n1 , n2 , …, nn ) dimensional n-space WW n , then U has a n-cyclic n-vector if and only if ordered n-basis for W in which U is represe

companion n-matrix of the n-minimal polynomia

Proof: We have just noted that if U has a n-cycl

there is such an n-ordered n-basis for W = W

Conversely if there is some n-ordered n-basis

… n

n n

1 k { , . . ., }D D for W in which U is repres

companion n-matrix of its n-minimal polynomi

that 1 nD D is a n-cyclic vector for U

Proof: One way to see this is to let U = U1 … U

operator on 1 nk k

1 nF . . . F which is represented by

… An in the n-ordered n-basis and to apply

theorem and the Cayley Hamilton theorem for n-vectYet another method is by a direct calculation.

Now we proceed on to define the notion o



p

decomposition or we can call it as cyclic n-decomp

its n-rational form or equivalently rational n-form.

Our main aim here is to prove that any n-linearon a finite (n1, …, nn) dimensional n-vector space V

Vn, there exits n-set of n-vectors 1

1{ , . .D

n

n n

1 r{ , . . ., }D D in V such that V = V1 … Vn

=1

1 1

1 1 r 1Z( ; T ) . . . Z( ; T )D D 2

1 2Z( ; T ) . . . ZD

… n

n n

1 n r nZ( ; T ) . . . Z( ; T ).D D

In other words we want to prove that V is a n-di

n-T-cyclic n-subspaces. This will show that T is a n

of a n-finite number of n-linear operators each of whcyclic n-vector. The effect of this will be to re

problems about the general n-linear operator to simil

about an n-linear operator which has a n-cyclic n-vec

This n-cyclic n-decomposition theorem is closel

the problem in which T n-invariant n-subspaces W

property that there exists a T- n-invariant n-subspac

that V = W W', i.e., V = V1 … Vn = W1 W



Thus n-admissibility characterizes those n-in subspaces which have n-complementary n- subspaces.

We see that the n-admissibility property is involve

decomposition of the n-vector space V = Z(D1; T) T); i.e., V = V1 … Vn =

1

1 1

1 1 r{Z( ; T ) . . . Z( ;D D



n

n n

1 n r n{Z( ; T ) . . . Z( ; T )}D D .

We arrive at such a n-decomposition by selecting n-vectors

1

1 1

1 r{ , . . ., }D D … n

n 1

1 r{ , . . .,D D

Suppose that by some method or another we ha

the n-vectors1 n

1 1 n n

1 r 1 r{ , . . ., } . . . { , . . ., }D D D D and n

which is proper say

W j =1 n

j jW ... W =

1

1 1

1 1 j 1{Z( ; T ) . . . Z( ; T )}D D …

n

n n

1 n j n{Z( ; T ) . . . Z( ; T )}D D .

We find the non zero n-vector1 n

1 n

j 1 j 1( . . . ) D D su

j j 1W Z( , T) 0 . . . 0D i.e.,

1 n

1 1 n n

j j 1 1 j j 1 n(W Z( ; T )) . . . (W Z( ; T D D

= 0 … 0

because the n-subspace.

j 1 j j 1W W Z( , T) Di.e.,

1 nW W W

If 1 n

1 1 n n

1 j 1 j{ , . . ., } . . . { , . . ., }D D D D have bee

W j is T-n-admissible n-subspace then it is rathe

suitablen

1 n

j 1 j 1. . . D D .

Let W = W1 … Wn be a proper T

subspace. Let us find a non zero n-vector D =

such that W Z(D; T) = {0} … {0}, i.e.,

W Z( T ) {0} {0}



… Wn Z(Dn; Tn) = {0} … {0}.

some n-vector E = E1 … En which is not in W

Wn i.e., each Ei is not in Wi; i = 1, 2 ,…, n. Coconductor S(E; W) = S(E1; W1) … S(En; Wn

of all n-polynomials g = g1 … gn such that g

… gn(Tn)En is in W = W1 … Wn. Re

monic polynomial f = S(E; W); i.e., f 1 …… S(En; Wn) which n-generate the n-ideals

W1) …S(En; Wn); i.e., each f i = S(EI; Wi) ge

S(EI; Wi) for i = 1, 2, …, n, i.e., S(E; W) i

conductor of E into W. The n-vector f(T)E = f

f n(Tn)En is in W = W1 … Wn. Now if W is

there is a J= J…Jn in W with f(T)

DEJand let g be any n-polynomial. Sinceg(T)Ewill be in W if and only if g(T)D is in W

S(DW) = S(EW). Thus the n-polynomial f

conductor of D into W. But f(T)D = 0 … f 1(T1)D … f n(Tn)Dn = 0 … 0; i.e., g

and only if g(T)D = 0 … 0 i.e., g1(T1)D1 0 … 0. The n-subspaces Z(D; T) = Z(D; T1

Tn) and W = W1 … Wn are n-independent a

ihil f

exists non zero n-vectors ^ `1

1 1

1 ! !r D D ^ 1 !n

rD D

respective T-n-annihilators ^ `1

1 1

1 ! r p p … ^ 1 !n p

that

(i) V = W o Z( D 1; T) … Z( D r ; T) =

^ `1 1 1

1 1 1; ; "o rW Z T Z T D D



^ `11 1 1; ;o r

^ `2

2 2 2

1 2 2; ; "o r W Z T Z T D D …

^ `1 ; ; "n

n n n

o n r nW Z T Z T D D .

(ii)t

t

k p divides 1t

t

k p , k = 1, 2,…, r; t = 1, 2, … ,

Further more the integer r and the n-annihilators

…^ `1 ,...,n

n n

r p p are uniquely determined by (i)

infact that not

t

k D is zero, t = 1, 2, …, n.

Proof: The proof is rather very long, hence they are four steps. For the first reading it may seem easier to

{0} …{0} =1

0W … n

0W ; i.e., eachi

0W =

1, 2, …, n, although it does not produce any

simplification. Throughout the proof we shall abbre

to f E. i.e., f 1(T1)E1 … f n(Tn)En to f 1E1 … f nESTEP I:

^ `n

n n n

0 1 n r nW Z ;T ... Z ;T E E

2. if 1 d k i d ri; i = 1, 2, …, n andWk =

n

k k nW W "

1

1

= ^ `1

1 1 1

o 1 1 k 1W Z ;T Z ;T E E"

^ `2 2 2



^ `2

2 2 2

o 1 2 k 2W Z ;T Z ;T E E"

^ `n

n n no 1 n k nW Z ;T Z ;T E E"

then the n-conductor

pk =1 n

1 n

k k p p !

= S

1 1

1 1

k k 1

;W

E …S

n n

n n

k k 1

;W

E

has maximum n-degree among all T-n-conduc

subspace Wk–1 = 1 n

1 n

k 1 k 1W W ! that is fo

k n);

n-deg pk = 1

1

1

1 1

k 1in V

max deg S ;W D

D …

n

nn

n n

k 1in V

max deg S ;W D

D .

This step depends only upon the fac

1 n

0 0W W ! is an n-invariant n-subspace. I

Wn is a proper T-n-invariant n-subspace then

0 < max n deg S ;W n dim VD d



(1, 1, …, 1) apply the n-division algorithms gi =

n-deg ri < n-deg f i.e.,

1 n 1 1

1 n 1 1

i i i i ig g f h r " … nf h

ri = (0 … 0) if n-deg ri < n-deg f.

We wish to show that ri = (0 … 0) for each

Letk 1



J =k 1

i i

1

h

E E¦ ;

i.e.,

J…Jn=1

1 1

k 11 1

1 i i

1

h§ ·

E E ¨ ¸© ¹

¦ … n

§ E ¨

©

Since J – E is in Wk–1 i.e., (J1 – E1) … (Jn – …

n

n

k 1W . Since

S i i

i i

i k 1 i k 1;W S ;W J E

i.e.,

1 n

1 n

1 k 1 n k 1S ;W S ;W J J!

= 1 n

1 n

1 k 1 n k 1S ;W S ;W E E"

= f 1 …f n.

S(J; Wk–1) = S(E; Wk–1) = f.

Further more f J =k 1

0 i i

1

E J E¦ i.e.,

§ · §

Wk–1 =1 n

1 n

k 1 k 1W W !

contains

W j–1 =1 n

1 n

j 1 j 1W W ! ;

the n-conductor f = S(J; Wk-1) i.e., f 1 … f n = S

… S(Jn;n

n

k 1W ) must n-divide p. p = fg i.e., p1

f 1g1 … f ngn. Apply g(T) = g1(T1) … gn(



sides i.e., pJ = gf J = gr j E j + gEo + i i

1 i j

gr

d

E¦ i.e.,

p1J1 … pnJn = g1f 1J1 … gnf nJn

=1 1 1 1

1 1

1 1 1 1 1

1 j j 1 0 1 i i

1 i j

g g g rd

§ ·J E E E¨ ¸¨ ¸

© ¹¦

2 2 2 2

2 2

2 2 2 2 2

2 j j 2 0 2 i i

1 i j

g g g rd

§ ·J E E E¨ ¸¨ ¸© ¹

¦ …

n n n n

n n

n n n n n

n j j n 0 n j j

1 i j

g g g rd

§ ·J E E E¨ ¸¨ ¸

© ¹¦ .

By definition, pJ is in W j–1 and the last two terms

side of the above equation are in W j-1 =1

1

j 1W

Therefore

gr jE j =1 1 n n1 j j n j jg r g rE E!

is in

W j-1 =1 n

1 n

j 1 j 1W W ! .

Now using condition (2) of step 1

= deg (S(J;1

1

j 1W )) … deg(S(Jn; W

= n-deg p

= deg p1 …deg pn

= n-deg fg= deg f 1g1 …deg f ngn.

Thus n-deg r j > n-deg f; i.e., deg1 jr … deg r



deg f n and that contradicts the choice of j =

now know that f = f 1 … f n, n-divides each

n

n

ig i.e.,ti

f dividest

t

ig for t = 1, 2, …, n and he

i.e., 1 n

0 0...E E = f 1J1 … f nJn. Since W0 = W

T-n-admissible (i.e., each k

0W is Tk -admissible f

n); we have E0 = f J0 where J

1 n

0 0...J J n

0W ; i.e., 1 n

0 0...E E = 1 n

1 0 n 0f ... f J J whe

make a mention that step 2 is a stronger form

that each of the n-subspaces W1 =1 n

1 1W ... W

n

2W , …, Wr =1 n

r rW ... W is T-n admissibl

STEP 3:

There exists non zero n-vectors (1

1D , …,1

1

rD )

n

n

rD ) in V = V1 … Vn which satisfy conditio

the theorem. Start with n-vectors {1

1 1

1 r, ,E E! }

n

n

rE } as in step 1. Fix k = (k 1, …, k n) as 1 d k i d r

2 to the n vector E = E E = 1E

where J0 = 1 n

0 0J J! is in W0 = 1

0W !

^ `1

1 1

1 k 1h , , h ! … ^ `n

n n

1 k 1h , , h ! are n-polynom

Dk = k 0 i i1 i k

hd E J E¦i.e.,

^ `1 n

1 n

k k D D! =

§ · §



1 1 1

1 1

1 1 1 1

k 0 i i

1 i k

h

d

§ ·E J E ¨ ¸¨ ¸

© ¹¦ !

n

n n

n n

k 0

1 i k

h

d

§ E J ¨ ¨

© ¦

Since Ek – Dk = 1 1 n n

1 1 n n

k k k k E D E D! is in

Wk–1 =1 n

1 n

k 1 k 1W W ! ;

S(Dk ;Wk–1) = S(Ek ;Wk–1)

= pk

= 1 1 n n

1 1 n n

k k 1 k k 1S , W S , W D D!

= 1 1 n n

1 1 n n

k k 1 k k 1S ,W S , W E E!

=1 n

1 n

k k p p !

and sincepk Dk = 0 …0.

1 1 n n

1 1 n n

k k k k p pD D! = 0 … 0.

we have

Wk–1 Z(Dk ; T) = {0} … {0}.

That is

1 1

1 1

k 1 k 1W Z ;T ... D 1 n

n n

k 1 k nW Z ;T D

^ `2

2 2 2

0 1 2 k 2W Z ;T ... Z ;T D D

^ `n

n n n

0 1 n k nW Z ;T ... Z ;T D D

and that pk =

1 n

1 n

k k p ... p is the T-n-annihilato

…n

n

k D . In other words, n-vectors^ 1

1 1

1 r, ,D D!

^ `n n d fi h



…, ^ `n

n n

1 r,...,D D define the same n-sequence

W1 = ^ `1 n1 1W ... W , W2 = ^ `1 n

2 2W ... W

vectors ^ ` ^ ` ̂ 1 2

1 1 2 2 n

1 r 1 r 1... , ... , , ..E E E E E !

T-n-conductors pk = S (Dk ; Wk–1) that is

^ `1 n

1 n

k k p ... p 1 1

1 1

k k 1S ;W S D D!

have the same maximality properties.The n-vect

^ `n

n n

1 r, ,D D! have the additional property that the

W0 = ^ `1 n

0 0W ... W ,

Z(D1; T) = 1 n1 1 1Z ;T ... Z ;TD D

Z(D2; T) = 1 n

2 2 2 nZ ,T ... Z ,TD D

are n-independent. It is therefore easy to verify

the theorem. Since

piDi = 1 1 n n

1 1 n n

i i i ip ... pD D = (0 …

h th t i i l l ti

=1 n

1 n

k k D D! ; pk n-divides each pi, i < k that is (

(k 1, …, k n) i.e.,1 n

1 n

k k p ... p n-divides each1

1

ip

each t

t

k p divides t

t

ip for t = 1, 2, …, n.

STEP 4:

The number r = (r1, …, rn) and the n-polynomials

(p p ) are uniquely determined by the co



…, (p1, …,nr

p ) are uniquely determined by the co

the theorem. Suppose that in addition to the^ ` ^ ` ^ `

1 2 n

1 1 2 2 n n

1 r 1 r 1 r,..., , ,..., ,..., ,...,D D D D D D we have no

vectors ^ ` ̂ `1 n

1 1 n n

1 s 1 s,..., ,..., ,...,J J J J with respec

annihilators ^ ` ̂ `1 n

1 1 n n

1 s 1 sg ,...,g ,..., g ,...,g such that

V = 0 1 sW Z ;T ... Z ;T J Ji.e.,

V = V1 … Vn

^ `1

1 1 1

0 1 1 s 1W Z ;T ... Z ;T J J …

^ `n

n n n

0 1 n s nW Z ;T ... Z ;T J J

t

t

k g dividest

t

k 1g for t = 1, 2, …, n and k t = 2, …,

show that r = s i.e., (r1, …, rn) = (s1, …, sn) and t

ip

n. i.e., 1 n

i ip ... p = 1 n

i ig ... g for each i. We see

= 1 n

1 1p ... p = 1 n

1 1g ... g . The n-polynomial g1 =

is determined by the above equation as the T-n-con

into Wo i.e., V = V1 … Vn into 1

0 0W ... W



hand fW for the set of all n-vectors f D = f 1D1 …D = D1 … Dn in W = W1 … Wn.

The three facts can be proved by the reader:

1. fZ(D; T) = Z(f D; T) i.e., f 1Z(D1; T1) …= 1 1 1 n n nZ f ;T ... Z f ;TD D .

2. If V = V1 … Vk ,

i.e., 1 1

1 kV ... V ... n n

1 kV ... V .



i.e.,11 k V ... V ...

n1 k V ... V .

where each Vt is n-invariant under T that is invariant under Ti ; 1 d i < t; t = 1, 2, …, n

fV1 … fVn; i.e., f 1V1 …f nVn = f 1

f 11

1

k V …n

n n

n 1 n k f V ... f V .

3. If D = D1 … Dn and J = J1 … Jn ha

T-n-annihilator then f D and f J have the annihilator and hence n-dim Z(f D; T) = n-di

i.e., f D = f 1D1 …f nDn and f J = f 1J1 …dim Z(f 1D1; T1) … dim Z(f nDn; Tn) =

T1) … dim Z(f nJn; Tn).

Now we proceed by induction to show that r = sfor i = 1, 2, …, r. The argument consists of c

dimensions in the proper way. We shall give the pro

r2, …, rn) t (2, 2,…, 2) then p2 = 1 n

2 2p ... p = g2 = g

and from that the induction should be clear. Suppo

(r1 … r

n) t (2, 2, …, 2); then n-dim W

0+ n-dim

n-dim V i.e., (dim 1

0W … dim n

0W ) + dim Z 1

1D

dim 1

0W + dim Z 1

1 1;TJ …dim n

0W + dim

dim V1 …dim Vn,

which shows that s = (s1, s2, …, sn) > (2, 2, …, 2

sense to ask whether or not p2 = g2,1 n

2 2p ... p

From the two decompositions, of V = V1 …two decomposition of the n-subspace



p2V =1 n

2 1 2 np V ... p V p2V = 1

2 0 2 1p W Z p ;T D

i.e.,1 n

2 1 2 np V ... p V =

1 1 1 1

2 0 2 1 1p W Z p ;T ... D n n

2 0p W Z p

2 2 0 2 1p V p W Z p ;T ... J 2Z p J

i.e.,1 n

2 1 2 np V ... p V = 1 1 1 1

2 0 2 1 1p W Z p ;T ... J

n n n n n

2 0 2 1 n 2... p W Z p ;T ... Z p J J

We have made use of facts (1) and (2) above an

the fact1 n

1 1 n n

2 i 2 i 2 ip p ... pD D D = 0 …0;

(2, 2, …, 2). Since we know that p1 = g1 fact (

that

1 1 n

2 i 2 1 1 2Z p ;T Z p ;T ... Z pD D Dand 1 1 n n

2 1 2 1 1 2 1 nZ p ;T Z p ;T ... Z p ;TJ J J ,

and g2 n-divides p2 i.e., t

2g divides t

2p for each t; t =

The argument can be reversed to show that p2 n-divt

2p divides t

2g for each t; t = 1, 2, …, n. Hence p2 = g

We leave the two corollaries for the reader to prove.

COROLLARY 1.2.17: If T = T 1 … T n is a n-line

on a finite (n1 , …, nn ) dimensional n-vector space V



f ( ) pV n then T-n-admissible n-subspace has a comple

subspace which is also invariant under T.

COROLLARY 1.2.18: Let T = T 1… T n be operator on a finite (n1 , …, nn ) dimensional n-vecto

V 1 … V n.

a. There exists n-vectors D = D 1 … D n… V n such that the T-n-annihilator of minimal polynomial for T.

b. T has a n-cyclic n-vector if and onlycharacteristic and n-minimal polynomial

identical.

Now we proceed on to prove the Generalized Cayle

theorem for n-vector spaces of finite n-dimension.

THEOREM 1.2.58:(G ENERALIZED C AYLEY H AMILTON

Let T = T 1 … T n be a n-linear operator on a fi…, nn ) dimension n-vector space V = V 1 …V n.be the n-minimal and n-characteristic polynom

iii. If p = 1

1 ... k r r

k f f is the prime factorizatio

1

1 ... k d d

k f f where d i is the n-multip

n-divided by the n-degree of f i.

That is if p = p1 … pn

= 11

1 11

1

1 1

1 1... ... ... n

k

n

r r r n n

k k f f f f

then f = 11

1 111 1 nnk k n

d d d d n nf f f f ;



then f 11 1... ... ...

nk k f f f f ;

is the nullity of ( )t

ir t

i t f T which is n-divided by t

i.e., t

id ; 1d i d k t . This is true for each t = 1, 2, …

Proof: The trivial case V = {0} … {0} is ob

(i) and (ii) consider a n-cyclic decomposition

V = Z(D1; T) … Z(Dr; T)

= 1

1 1

1 1 r 1Z ;T ... Z ;TD D …

n

n n

1 n r nZ ;T ... Z ;TD D .

By the second corollary p1 = p. Let U1 = 1

iU !

restriction of T = T1 …Tn i.e., each s

iU is t

Ts (for s = 1, 2, …, rs) to s

i sZ ;TD . Then Ui h

vector and so pi =1 n

1 n

i ip p ! is both n-mini

and the n-characteristic polynomial for Ui. T

characteristic polynomial f = f 1 …f n is th

Let p = 1 n1 hk k 1 11 n

1 n

r rr r1 1 n n

1 k 1 kf f f f ! ! ! b

n-factorization of p. We employ the n-primary dec

theorem, which tell us if

i i i

t 1 nV V V !

is the nfor

tiri

t tf T then

V = V1 … Vk = 1k 1

1 1V V ! ! 1

nV

and tirt

if is the minimal polynomial of the o



and if is the minimal polynomial of the o

restricting Tt to the invariant subspace itV . This is tru

= 1, 2, …, n. Apply part (ii) of the present theore

operator i

tT . Since its minimal polynomial is a po

prime t

if the characteristic polynomial for i

tT ha

tirt

if where

t t

i id r! , t = 1, 2, …, n.

We havet

id =i

t

t

i

dim V

degf for every t = 1, 2 ,…, n a

= nullity tirt

i tf T for every t = 1, 2, …, n. Since Tt

sum of the operators tk 1

t tT , ,T! the characteristic po

is the product, f t = ttk i t

t

ddt t

1 k f f " . Hence the claim

The immediate corollary of this theorem is left as

for the reader.

COROLLARY 1.2.19: If T = T 1 … T n is a operator of the n-vector space of (n1 n ) dimens

for Z( D i; T) = Z 1

1

1;i T D … ;n

n

i n Z T D . H

denotes n-dimension of Z( D i; T) that is the n-d

annihilator pi =1

1...

n

n

i i

p p . The n-matrix

operator T i in the ordered n-basis Bi is the nmatrix of the n-polynomial pi. Thus if we let ordered basis for V which is the n-union of Bi ar

^ `1

1 1

1 , ,! r B B …^ `1 , ,!n

n n

r B B then the n-ma



ordered n-basis B will be A = A

1

A

2

… A

n

=

1

1

1

1

2

1

r

A 0 0

0 A 0

0 0 A

ª º« »« »« »« »

« »¬ ¼

"

"

# # #

"

…

n

1

n

2

A 0

0 A

0 0

ª « « « «

« ¬

# #

where t

iA is the t t

i ik k u companion matrix of

…, n. A (n1 u n1, …, nn u nn) n-matrix A which

sum of the n-companion matrices of the non-s

polynomial^ `1

1 1

i rp , ,p! …^ `n

n n

1 rp , ,p! such t

t

t

ip for it = 1, 2 ,… , rt – 1 and t = 1, 2, …, n w

the rational n-form or equivalently n-rational for

THEOREM 1.2.59: Let F = F 1 … F n be a n-f

… Bn be a (n1 u n1 , …, nn u nn ) n-matrix ovn-similar over the n-field F to one and only on

rational form

similar over the field to one and only one n-matrix C

the rational n-form.

The n-polynomials ^ 1 n

1 1 n n

1 r 1 rp , ,p , , p , ,p! ! !

invariant n-factors or n-invariant factors for the n-

B1… Bn.We shall just introduce the notion of n-Jord

Jordan n-form. Suppose that N = N1 …Nn be a

linear operator on a finite (n1, n2,…, nn) dimension sp



… Vn. Let us look at the n-cyclic decompos

which we have depicted in the theorem. We ha

integers (r1, …, rn) and non zero n-vectors ^ n

1 ,D !

with n-annihilators ^ ` ^ `1 n

1 1 n n

1 r 1 rp , , p p , , p ! ! !

V = 1 rZ ; N Z ;ND D"

= 1

1 1

1 1 r 1Z ; N Z ; ND D " "

n

n n

1 n r nZ ;N Z ; ND D"

andt

t

i 1p dividest

t

ip for it = 1, 2 ,…, rt – 1 and t =

Since N is n-nilpotent the n-minimal poly1 nk k

x x ! with k t d nt; t = 1, 2, …, n. Thus ea

the formti

t

k t

ip x and the n-divisibility condition

t

t t t

1 2 rk k k t t t! ; t = 1, 2, …, n. Of courset

1k = k t

The n-companion n-matrix of 1 n

i i1 nk k

x x ! is the

matrix A =1 n

A A with

Thus we from earlier results have an n-ordered

V1 … Vn in which the n-matrix of N is the

the elementary nilpotent n-matrices the size

decrease as itincreases. One sees from this that

a n-nilpotent (n1 un1, …, nn unn), n-matrix is a

integers (r1, …, rn) i.e.,

^ `1

1 1

1 rk , , k ! …^ `n

n n

1 rk , ,k !

such that1 nk k



1

1 n

1 r 1k k n !

2

2 n

1 r 2k k n !

and

n

n n

1 r nk k n !

andt t

t t

i i 1k k t ; t = 1, 2, …, n and 1 d i, i + 1 d rt

of positive integers determine the n-rational matrix, i.e., they determine the n-matrix up to sim

Here is one thing, we like to mention about

n-operator N above. The positive n-integer

precisely the n-nullity of N infact the n-null spa

with (r1, …, rn) n-vectors

i i1 n

1

k 1 k 11

1 i nN N

D !

D1 …Dn be in n-nullspace of N we write

D= 1 1

1 1 1 1

1 1 r rf f D D! … n n

1 1f D !

where 1 n

1 n

i if , ,f ! is a n-polynomial the n- degr

may assume is less than1 n

i ik , , k ! . Since ND

i.e., N1D… NnDn = 0 …0 for each ir w



t

t

iN be the n-linear operator ont

t

iW defined by

1 d it d k t; thent

t

iN is n-nilpotent and has n-mini

tit

t

r

ix . On t

t

iW , Tt acts like t

t

iN plus the scalaidentity operator. Suppose we choose a n-ba

subspace1 n

1 n

i iW W ! corresponding to

decomposition for the n-nilpotentt

t

iN . Then the

this ordered n-basis will be the n-direct sum of n



this ordered n basis will be the n direct sum of n

1 2

1 2

1 2

1

c 0 ... 0 0 c 0 ... 0

1 c 0 0 1 c 0

c c

0 0 1 c 0 0 1 c

ª º ª « » « « » « « » « « » « « » « « » « ¬ ¼ ¬

# # # # # # #

! !

n

n

n

n

c 0 ... 0 0

1 c 0 0

c

0 0 1 c

ª º« »« »

« »« »« »« »¬ ¼

!

# # # #

!

each with c =t

t

ic for t = 1, 2, …, n. Further m

these n-matrices will decrease as one reads fromn-matrix of the form described above is called a

d i i h h i i l

A =

1

1 n

1 1

1 n

2 2

1

k

A 0 0 A 0

0 A 0 0 A

0 0 A 0 0

ª º ª « » « « » « « » «

« » « « » « ¬ ¼ ¬

! !

! "!

# # # # #

! "

of the k i sets of n-matrices ^ `1

1 1

1 k A , ..., A …^ AEach



t

t

t

t

i

t1

i

t t 2i

i

tn

J 0 ... 0

0 J ... 0A

0 0 ... J

ª º« »« »

« »« »« »¬ ¼

#

where each ti

jtJ is an elementary Jordan m

characteristic valuet

t

ic ; 1 < it < k t; t = 1, 2, …, n. A

eacht

t

iA the sizes of the matrices t

t

i

tjJ decrease as jt i

jt d nt ; t = 1, 2, …, n. A (n1 u n1, …, nn u nn) n-matrsatisfies all the conditions described so far for som

distinct k i-scalars ^ `1

1 1

1 k c c! …^ `n

n n

1 k c c! wil

be in Jordan n-form or n-Jordan form.

The interested reader can derive several

properties in this direction.

In the next chapter we move onto define the n

inner product spaces for n-vector spaces of type II.

Chapter Two



n-INNER PRODUCT

SPACES OF TYPE II

Throughout this chapter we denote by V = V1 n-vector space of type II over n-field F = F1

notion of n-inner product spaces for n-vector spa

introduced. We using these n-inner product

notion of n-best approximations n-orthogo

quadratic n-form and discuss their properties.

DEFINITION 2.1: Let F = F 1 … F n be a

a. ( DEJ ) = ( DJ ) + ( EJ )

i.e., ( D 1+ E 1 / J 1 ) …( D n+ E n / J n ) =

1 1 1 1 / / / /ª º ª ¬ ¼ ¬ " n n nD J E J D J E

b. (cD / E ) = (cD / E ) i.e.,(c1D 1 / E 1 ) … (cnD n / E n )

= c1( D 1 / E 1 ) … cn( D n / E n )

c. ( D / E ) = ( E / D ) i.e., ( D 1 / E 1 ) … ( D n / E n )

= ( E 1 / D 1 ) … ( E n / D n )

d (D / D ) = (D / D ) (D /D ) > (0



d. ( D / D ) = ( D 1 / D 1 ) … ( D n / D n ) > (0

D i z 0 for i = 1, 2, …, n.

From the above conditions we have

( D / c E + J ) = c ( D / E ) + ( D / J )

= ( D 1 / c1 E 1 + J 1 ) … ( D n / cn E n + J n )

= [c1 ( D 1 / E 1 ) + ( D 1 / J 1 )] … [cn ( D n / E n )

A n-vector space V = V 1 … V n endowed with

product is called the n-inner product space. Let F =

F n for V = 1

1 ... nnn

n F F a n-vector space over F th

inner product called the n-standard inner product. I

for D = D 1 …D n = 1

1 1

1 ! n x x … 1 !n

n n

n x x

and E = E 1 …E n = 1

1 1

1 ! n y y … 1 !n

n n

n y y

1 1

1

1 1 / ¦ ¦!

n n

n

n n

j j j j

j j

x y x yD E .

If A = A1… An is a n-matrix over the n-field F = F n where

u i in n

i i A F is a vector space over F i for i =

THEOREM 2.1: If V = V 1 … V n is an n-inne

then for any n-vectors D = D 1 … D n , E = E 1

and any scalar c = c1… cn.

(1) || cD || = | c | ||D ||

i.e., || cD|| = || c1D || … ||cn D n||

= |c1| ||D 1|| … |cn| ||D



(2) ||D || > (0, …, 0) = (0 … 0) for Dz0 i.e., || D || …|| D n|| > (0, 0, …, 0) = (0

(3) || ( DE|| d || D || || E ||

|| ( D E || … || ( D n / E n ) ||

d ||D 1|| || E || … || D n|| || E n||.

Proof as in case of usual inner produce space.

We wish to give some notation for n-inner produ

J = E – 2

( / )

|| ||

E D

DD

i.e., J1 …Jn = 1 11 12

1

( / )

|| ||

§ ·E DE D¨ ¸

D© ¹

n nn n2

n

( / )|| ||§ ·E DE D¨ ¸D© ¹

.

i.e., ( D 1 / E 1 ) … ( D n / E n ) = 0 … 0.

Since this implies E = E 1… E n is n-orthog

D 1… D n.

We often simply say that D and E are n-orthogon

S1… Sn be n-set of n-vectors in V = V1…called an n-orthogonal n-set provided all n-pair of n

vectors in S are n-orthogonal. An n-orthonormal n-

orthogonal n-set with addition property || D || = || D



||Dn|| = 1 … 1 for every D = D1 …Dn in S =Sn.

THEOREM 2.2: A n-orthogonal n- set of non-zero n-vlinearly independent.

The proof is left as an exercise for the reader.

THEOREM 2.3: Let V = V 1 … V n be an n-inn space and let

^ ` ^ `1

1 1

1 1, , ... , , ! !n

n n

n n E E E E

be any n-independent vectors in V. Then one way to orthogonal vectors

^ ` ^ `1

1 1

1 1, , , , ! ! !n

n n

n nD D D D

in V = V 1 … V n is such that for each k = 1, 2, …

^ ` ^ `1

1 1

1 1, , , , ! ! !n

n n

k k D D D D

is a n-basis for the n-subspace spanned by

^ ` ^ `1 1

1 1, , , , ! ! !n n

k kE E E E .

Suppose ^ ` ^ `1 n

1 1 n n

1 m 1 m, , , ,D D D D! ! ! (1 d

…, n) have been chosen so that f

^ ` ^ `1 n

1 1 n n

1 k 1 k , , , ,D D D D! ! ! ; 1 d k i d mi, i

an n-orthogonal basis for the n-subspace of V =

that is spanned by ^ ` ^1

1 1 n

1 k 1, ,E E E! !

construct next n-set of n-vectors,1 n

1 n

m 1 m 1, , D D!



let 1

1 1

1 1

1 1

1 1mm 1 k 1 1 1

m 1 m 1 k1 2k 1 k || ||

E DD E D

D¦

and so on

nn n

n n

n n

n nmm 1 k n n n

m 1 m 1 kn 2k 1 k || ||

E DD E DD¦

Thus 1 n 1 n

1 n 1 n

m 1 m 1 j j| D D D D! ! = (0

jt d mt; t = 1, 2, …, n. For other wise n

1

m 1 E be a n-linear combination of ^ `

1

1 1

1 m, , ...D D !

Further 1 d jt d mt; t = 1, 2, …, n.

Then

(Dm+1 / D j) =

1 1 n n

1 1 n n

m 1 j m 1 j D D D D!

1

1 1

1 1mm 1 k 1 1 1

ª E D« E ¦



a n-vector in V. To find the n-best approxima

… E n by n-vectors in W = W 1 … W n find a vector D = D 1 … D n for which || E

D 1|| … ||E n – D n|| is as small as possiblerestriction that D = D 1 … D n should belo

… W n. i.e., to be more precise.

A n-best approximation to E = E 1 …

… W n is a n-vector D = D 1 … D n in W s

|| ED || < || EJ ||



i.e.,|| E D || …|| E nD n || d || E 1J 1 || …

for every n-vector J = J 1 … J n in W.

The following theorem is left as an exercise for t

THEOREM 2.4: Let W = W 1 … W n be a n-suinner product space V = V 1 … V n and E =

a n-vector in V = V 1 … V n.

a. The n-vector D = D 1 … D n in W

approximation to E = E 1 … E n by n-vec

… W n if and only if E – D = E 1 – D 1

n-orthogonal to every vector in W. i.e., eorthogonal to every vector in W i; true for i =

b. If a n-best approximation to E = E 1 …

in W = W 1 … W n exists, it is unique.

c. If W is finite dimensional and 1

1 , ,!D

1 , ,!n

n n

nD D is any n-orthonormal n-basis

DEFINITION 2.4: Let V = V 1 … V n be n-inn

space and S = S 1 … S n any set of n-vectors i

orthogonal complement of S denoted by S A = S 1A

the set of all n-vectors in V which are n-orthogonalvector in S.

Whenever the n-vector D = D1 … Dn in the t

exists its is called the n-orthogonal projection to E =



En on W = W1 … Wn. If every n-vector has an n projection on W = W1 … Wn the n-mapping tha

each n-vector in V its n-orthogonal projection o

… Wn is called the n-orthogonal projection of V

We leave the proof of this simple corollary as an e

the reader.

COROLLARY 2.2: Let V = V 1 … V n be an n-in

space, W = W 1 … W n a finite dimensional n-su

E = E 1 … E n the n-orthogonal projection of V

the n-mapping E o E – E E i.e., E 1 …E n o ( E

… ( E n – E n E n ) is the n-orthogonal projection of V o

The following theorem is also direct and can be pro

reader.

THEOREM 2.5: Let W be a finite dimensional n-sub

n-inner product space V = V 1 … V n and let E = E n be the n-orthogonal projection of V on W. Then E

idempotent n-linear transformation of V onto space W.

COROLLARY 2.4: Let 1

1 1

1

{{ , , }n

D D ! 2

1

{ ,D !

1{ , , }}n

n n

nD D ! be an n-orthogonal set of nonze

an n-inner product space V = V 1 … V n. If Eis any n-vector in V then

1 2 2 2



1 2

1 21 2

1 21 2 2 2| ( | ) | | ( | ) |

|| || || || K K

K K K K

E D E D D D ¦ ¦

2

2

12

| ( | ) ||| || ||

|| ||

n

n n

n

n K

nn K K

E D E E

D d ¦ !

and equality holds if and only if

1

1

1 1

1

1 1

2

( | )

|| ||

K

K

K K

E D E D

D ¦ !

12

( | )

|| ||

n

n

n

n K n

K nn

Kn

E D D E E

D

¦ !

Now we proceed onto define the new n

functionals and adjoints in case of n-vector spac

of type II.

Let V = V1 … Vn be a n-vector space ov

(T11 | 1) … (Tnn | n) = (1 | *

1T 1) … (n

all , in V.

By the use of an n-orthonormal n-basis th

operation on n-linear operators (passing from Tidentified with the operation of forming the

transpose of a n-matrix.

Suppose V = V1 ! Vn is an n-inner pro

over the n-field F = F1 ! Fn and let = 1 some fixed n-vector in V. We define a n-function f

h l fi ld b f ( ) ( |)



the scalar n-field by f ( )E D = (|);

i.e.11f E (1) …

nnf E (n) = (1 | 1) … (n |

for = 1 … n V = V1 … Vn.

The n-function f is a n-linear functional on V bec

very definition (|) is a n-linear n-function on V, away from some = 1 … n V.

THEOREM 2.6: Let V be a finite (n1 , …, nn ) dimensio

product space and f = f 1 … f n , a n-linear functio

V 1 … V n. Then there exists a unique n-vector

n in V such that f() = ( | ) i.e. f 1(1 ) … f n(n … (n| n ) for all in V.

Proof: Let1 n

1 1 n n

1 n 1 n{ } { }D D D D! ! ! be a n-orth

basis for V. Put

= 1 … n

=1 nn n

1 1 n nf ( ) f ( )D D D D¦ ¦

=1 1

1 n

1 1 1 n

K 1 j j K

j j

| f ( ) | f§ · §

D D D D ¨ ¸ ̈ ¨ ¸ ̈ © ¹ ©

¦ ¦!

= f 11

K ( )D … f nn

K ( )D .

Since this is true for each K =1 n

K K D D! , it

(f ). Now suppose J = J1 … Jn is a n-vecto

(|) = (|J) for all V; i.e.,(1|1) … (

… (n | Jn). Then ( – J | – J) = 0 and =

tl t 1 d t i



exactly one n-vector = 1 … n determinfunctional f in the stated manner.

THEOREM 2.7: For any n-linear operator T = T

a finite (n1 , …, nn ) dimensional n-inner product

… V n , there exists a unique n-linear operator *

nT on V such that (T|) = ( | T * ) for all , i

Proof: Let = 1 … n be any n-vector i

(T|) is a n-linear functional on V. By the earl

is a unique n-vector c = 1cE … n

cE in V su

(|c) for every in V. Let T* denote the mappi

T*. We have (T1 1| 1) … (Tnn | n) = (

(n |*

nT n), so must verify that T* = *

1T …

linear operator. Let , J be in V and c = c1 scalar. Then for any ,

( | T* (c + J)) = (T | c + J)

The uniqueness of T* is clear. For any in V the n-v

uniquely determined as the vector c such that (T|)every .

THEOREM 2.8: Let V = V 1 … V n be a finite

dimensional n-inner product space and let B =1

1{D !

1{ }n

n n

nD D ! be an n-ordered n-orthogonal basis f

= T 1 … T n be a n-linear operator on V and let A

A b h f T h d d b B



An be the n-matrix of T in the ordered n-basis B. (T j| K ) i.e.

1 1

1

n n

n

K j K j A A ! =1 1

1 1

1( | ) ( |n

n

j K n jT T D D D !

Proof: Since B =1 n

1 1 n n

1 n 1 n{ } { }D D D D! ! ! is an

basis we have

D =1 n

1 n

1 n

n n1 1 n n

1 K K n K K

K 1 K 1

( | ) ( | )

D D D D D D¦ ¦!

= 1 … n.

The n-matrix A is defined byn

j Kj K

K 1

T A

D D¦ , i.e.,

T1

1 n

1 n 1 1 1

1 n

n n1 n 1 1 n

j n j K j K K

K 1 K 1

T A A

D D D ¦ ¦! !

sincen

j j K K T (T | )D D D D¦ ;

The following corollary is immediate and left f

prove.

COROLLARY 2.5: Let V be a finite (n1 , …, nn )inner product space over the n-field F and set Ta n-linear operator on V. In any n-orthogonal bamatrix of T * is the n-conjugate transpose of the n

DEFINITION 2.5: Let T = T 1 … T n be a n-

i d t V V V th



on an inner product space V = V 1 … V n , the

= T 1 … T n has an n-adjoint on V if there e

operator T *= *

1T … *

nT on V such that (T|

(T 11| 1 ) … (T nn | n ) = (1 | *

1T 1 ) …

all = 1 … n and = 1 … n in V =

It is left for the reader to prove the following theo

THEOREM 2.9: Let V = V 1 … V n be a finite

inner product space over the n-field F = F 1 …U are n-linear operators on V and c is n-scalar.

i. (T + U)* = T * + U *

i.e., (T 1 + U 1 )* … (T n + U n )

*

= * *

1 1T U … ( * *

n nT U ).

ii. (cT)* = cT *.

iii. (TU)*

= U *

T *

.iv. (T * )* = T.

THEOREM 2.10: Let V and W be finite dimensio

product spaces over the same n-field F = F 1 …

and W are of same n-dimension equal to (n1 ,…,nn ).

… T n is a n-linear transformation from V i following are equivalent.

i. T = T 1 … T n preserves n-inner products

ii. T = T 1 … T n is an n-isomorphism.

iii. T = T 1 … T n carries every n-orthonor

for V into an n orthonormal n basis for Wi T i h l b i f V



for V into an n-orthonormal n-basis for W.iv. T carries some n-orthonormal n-basis for V

orthonormal n-basis for W.

Proof: Clearly (i) o (ii) i.e., if T = T1 … Tn p

inner products, then |T|| = |||| for all = 1 …

i.e., || T11|| … || Tn n || = || 1 || … || n ||. Tnon singular and since n-dim V = n-dim W = (n1,

know that T = T1 … Tn is a n-vector space n-iso

(ii) o (iii) Suppose T = T1 … Tn is an n-isomo

1 n

1 1 n n

1 n 1 n{ } { }D D D D! ! ! be an n-orthonormal b

Since T = T1 … Tn is a n-vector space isomorp

dim V = n-dim W, it follows that 1

1

1{T ,D !

2

1 2

2 2 2 n{T , ,T }D D! … n

n n

n 1 n n{T , ,T }D D! is a n-b

Since T also n-preserves inner products (T j | TK ) =

G jK , i.e., (T1 1 1

1 1

j 1 K | T )D D … n n

1 n

n j n K (T | T )D D

i.e.,1 1 n n

1 1 n n

1 j 1 K n j n K {T |T } {T |T }D D D D!

=1 1 n n

1 1 n n

j K j K ( | ) ( | )D D D D!

=1 1 n n j K j K G G! .

For any

= 1 ! n

=1 1

1 1 1 1

1 1 n nx xD D ! ! n

n n n

1 1 nx xD !

and

E =1 1 n

1 1 1 1 n n n

1 1 n n 1 1 n

y y y yD D D D! ! !



in V we have

(|) =n

j j

j 1

x y

¦

that is

[

1

1 1

1

n

1 11 1 n n j j

j 1( | ) ( | ) ] x y

D E D E ¦! !

(T | T) = (T11 | T11) … (Tnn | Tnn)

= j j K K

j K

x T y T§ ·

D D¨ ¸¨ ¸© ¹

¦ ¦

=1 1 1 1

1 1

1 1 1 1

j 1 j K 1 K

j K

x T y T§ ·

D D ¨ ¸¨ ¸© ¹¦ ¦ !

n n n n

n n

n n n n

j n j K n K

j K

x T y T§ ·

D D¨ ¸¨ ¸© ¹

¦ ¦

=n

j j

j 1

x y

¦

=n

1 1 n n1 n

nn1 1 n n

j j j j

j 1 j 1

x y x y

¦ ¦!

and so T-n-preserves n-inner products.

COROLLARY 2.6: Let V = V 1 … V n and W = W 1be finite dimensional n-inner product spaces over

field F = F F Then V and W are n isomo



field F = F 1 … F n. Then V and W are n-isomoonly if they have the same n-dimension.

Proof: If 1 n

1 1 n n

1 n 1 n{ } { }D D D D! ! ! is a n-orth

basis for V = V1 … Vn and1

1 1

1 n{ } {E E ! !

an n-orthonormal n-basis for W; let T be th

transformation from V into W defined by T

1 n 1 n

1 n 1 n

1 j n j j jT TD D E E! ! . Then T is an n-is

of V onto W.

THEOREM 2.11: Let V = V 1 … V n and W = W 1be two n-inner product spaces over the same n-field

T 1 … T n be a n-linear transformation from V inT-n-preserves n-inner products if and only if ||T|every in V.

Proof: If T = T1 … Tn, n-preserves inner produn-preserves norms. Suppose ||T|| = || || for every

It is left for the reader to verify that the product o

operator is n-unitary.

THEOREM 2.12: Let U = U 1 … U n be a n-on an n-inner product space V = V 1 … Vunitary if and only if the n-adjoint U * of U exU *U = I.

Proof: Suppose U = U1 … Un is n-unitar

invertible and (U | ) = (U|UU-1) = ( | U-1



invertible and (U | ) (U|UU ) ( | U

… n and = 1 … n in V. Hence U-1

of U for

(U11 | 1) … (Un n | n)

= (U11 | U11

1U 1) … (UnDn |Un1

nU

= (1 |1

1U

1) ! (n |1

nU

n).

Conversely suppose U* exists and UU* = U*U =

U1*

1U … Un*

nU

= *

1U U1 … *

nU Un

= I1 … In.

Then U = U1 … Un is n-invertible wit1 1

1 nU U ! = *

1U … *

nU . So we need o

= U1 ! Un preserves n-inner products. W

= ( | U*U) = (|I) = ( | );

i.e.,

(U | U ) (U | U )

An to be n-anti orthogonal if AtA= – I, i.e., t

1A A1

= –I1 –I2 … –In.

Let V = V1 … Vn be a finite dimensio

product space and T = T1 … Tn be a n-linear opWe say T is n-normal if it n-commutes with its n

TT* = T*T, i.e., T1*

1T … Tn*

nT = *

1T T … T

The following theorem speaks about the properties

n-self adjoint n-operators on a n-vector space over th

type II.



yp

THEOREM 2.13: Let V = V 1 … V n be an n-in

space and T = T 1 … T n be a n-self adjoint opeThen each n-characteristic value of T is recharacteristic vectors of T associated with

characteristic values are n-orthogonal.

Proof: Suppose c = c1 … cn is a n-characteristic

i.e., T = c for some nonzero n-vector = 1 …

T11 … Tnm = c11 … cnn.

c(|) = (c|) = (T | )

= ( |T) = ( | c)

= c (|)

i.e.,

c(|) = c1 (1 |1) … cn(|n)

= (c1

1|

1) … (c

n

n|

n)

= (T11 | 1) … (Tnn|n)

( | T ) ( |T )

= ( | d) = d(| )

= d( | ).

If c z d then ( | ) = 0 … 0; i.e., if d1 …

cn i.e., ci z di for i = 1, 2, …, n then (1 | 1) 0 … 0.

The reader is advised to derive more properties i

We derive the spectral theorem for n-inner p

space over the n-real field F = F1 … Fn.



THEOREM 2.14: Let T = T 1 … T n be aoperator on a finite (n1 , …, nn ) dimensional

vector space V = V 1 … V n. Let

1

1 1

1 1{ } { }n

n n

K K c c c c ! ! !

be the n-distinct n-characteristic values

1

1

n

n

j j jW W W ! be a n-characteristic space

n-scalar 1

1

n

n

j j jc c c ! and 1

1

j j E E !

orthogonal projection of V on W j i.e., V t on t

jW

n. Then W i is n-orthogonal to W j if i z j1

1

n

n

i iW W ! and W j = 1

1

n

n

j jW W ! then W

tot

t

jW if it z jt for t = 1, 2, …, n. V is the n

1

1 1

1 1{ , , } { , , } ! ! !n

n n

K K W W W W and

T = T 1 … T n

= 1 11 11 1 1 1[ ] K K c E c E ! ! 1 1[ n nc E c !

Proof: Let = be a n vector in

Hence (c j – ci) (|) = 0 ! 0. i.e.,

1 1 n n

1 1 n n

j i 1 1 j i n n(c c )( | ) (c c )( | ) D E D E! = 0

since c j – ci z 0 … 0. It follows (|) = (1 | (n|n) = 0 … 0. Thus W j =

1 n

1 n

j jW W ! is n

to Wi =1 n

1 n

i iW W ! when i z j. From the fact th

n-orthonormal basis consisting of n-characteristic,

that V = W1 + … + WK i.e.,

V = V1 ! Vn 1 1



=

1 n

1 1 n n

1 K 1 K(W W ) (W W ) ! ! !

when it z jt ; 1 d it, jt d K t or t = 1, 2, …, n. From the

= V1 … Vn has an n-orthonormal n-basis cons

characteristic n-vectors it follows that V = W1 + …

t t

t t

j j t tW (1 j K )D d d and 1

1 1 n

1 K 1( ) (D D D ! ! = 0 … 0 then 0 … 0

= i j

j

( | )§ ·

D D¨ ¸© ¹

¦

= i j

j

( | )D D

¦

i.e.1 1 n n

1

1 1 n n

i j i j

j

( | ) ( | )§ · § ·

D D D D¨ ¸ ¨ ¸¨ ¸ © ¹© ¹¦ ¦!

1 1 n n

1 n

1 1 n n

i j i j

j j

( | ) ( | ) D D D D¦ ¦!

1 n

1 2 n 2i i|| || || || D D!

This decomposition is called the n-spectral resolu

following corollary is immediate however we sk

of it.

COROLLARY 2.7: If

1

1 !n

n

j j je e e

= 1

1 1 1 1

1

1 1z

§ ·¨ ¸¨ ¸

© ¹

i

i j j i

c

c c …

z

§ ¨ ¨

©

n n n

ni j j

x c

c

then E = 1 nE E for 1d j d K ; t = 1 2



© ¹ ©then E j =1

!n j j E E for 1d jt d K t ; t = 1, 2, …

Proof: Sincet t

t t

i jE E 0 for every 1 < it, jt < K t; t

z jt) it follows that T2 = (T1 … Tn)2 = 2

1T

1 1

1 2 1 1 2 1 n 2 n1 1 K K 1 1(c ) E (c ) E (c ) E ! ! !

and by an easy induction argument that

1 1

n 1 n 1 1 n 1

1 1 K K T (c ) E (c ) Eª º ¬ ¼! !

n n

n n n n n n

1 1 K K(c ) E (c ) Eª º ¬ ¼!

for every n-integer (n1, …, nn) > (0, 0, …, 0). For

polynomialr

n

n

n 0

f a x

¦ ;

f = f 1 … f n =

1 n

1

1

1 n

r r n1

n

n 0 n 0

a x a

¦ ¦!

we havef(T) = f 1(T1) … f n(Tn)

=1 1

1

1 1 1

1 1

K r n1 1

n j j

j 1 n 0

a c E

§ · ¨ ¸¨ ¸

© ¹¦ ¦ !

n n

n

n n

n n

K r nn

n j

j 1 n 0

a c

§ ·¨ ¨ © ¹

¦ ¦

=

1 n

1 1 n n

1 n

K K 1 1 n n

1 j j n j j j 1 j 1f (c )E f (c )E ¦ ¦

!.

Since1 1 1 1

1 t

j m j me (c ) G for t = 1, 2, …, n it follows e

true for t = 1, 2, …, n.

Since1 n

1 1 n n

1 K 1 K {E E } {E E } ! ! ! are canonically

with T and I1 I = ( 1 1 nE E ) (E



with T and I1 … In (11 K 1E E ) (E ! !

the family of n-projections1

1 1 n

1 K 1{E E } {E ! !

called the n-resolution of the n-identity defined by T

Tn.

Next we proceed onto define the notion of n-diago

the notion of n-diagonalizable normal n-operators.

DEFINITION 2.7: Let T = T 1 … T n be a n-dia

normal n-operator on a finite dimensional n-inn

space and

T = T 1 … T n =1

1 1

1

1 1

1 1

¦ ¦!n

n

n

K K n

j j j

j j

c E c E

is its spectral n-resolution.

Suppose f = f 1 … f n is a n-function whose n-

includes the n-spectrum of T that has n-values in the scalars. Then the n-linear operator f(T) = f 1(T 1 ) …

d f d b h

THEOREM 2.15: Let T = T 1 … T n be a n

normal n-operator with n-spectrum S = S 1 …

(n1 , …, nn ) dimensional n-inner product space

V n. Suppose f = f 1 … f n is a n-function

contains S that has n-values in the field of n-scis a n-diagonalizable normal n-operator with n-

f 1(S 1 ) … f n(S n ). If U = U 1 … U n is a n-u

onto V c and T c = UTU -1 = U 1T 11

1

! n nU U T

… S n is the n-spectrum of T c and f(T c ) =

f n(T c n ) = Uf(T)U –1 = U 1 f 1(T 1 ) 11 ( ! n n nU U f T



Proof: The n-normality of f(T) = f i(Ti) … f

a simple computation from the earlier results and

f(T)

*

= f 1(T1)

*

… f n(Tn)

*

=

1 1 n n

1 n

1 1 n n

j j j j

j j

f (c )E f (c )E ¦ ¦! .

Moreover it is clear that for every = 1 …

1 n

1 n

j 1 j nE (V ) E (V ) ! ; f(T) = f(c j), i.e., f 1(

f n(Tn)n = f 11

1

j(c ) 1 … f nn

n

j n(c )D . Thus th

f(c) with c in S is contained in the n-spectrum of

Conversely, suppose = 1 … n = 0

that f(T) = b i.e., f 1(T1)1 … f n(Tn)n = b1

Then = j

j

E D¦ i.e.,

1 n = 1

jE¦ 1 n

jE D¦

Hence2

j j

j

(f (c ) b)E D¦

=1 1

1

2

1 1

1 j 1 j 1

j

(f (c ) b ) E D¦ … n

n

n

n j n

j

(f (c ) b )E ¦

=1 1

1

2 21 1

1 j 1 j 1 j

f (c b ) E D

¦ …

nn

2n

n j n j

f (c b )

¦



¦ ¦Therefore f(c j) = f 1

1

1

j(c ) … f nn

n

j(c ) = f 1 …

… n

n

j nE D = 0 … 0; by assumption = 1

0 … 0 so there exists an n-index tuple i = (i1,

that Ei =1 ni 1 i nE ED D! = 0 … 0. By assu

1 … n z 0 … 0. It follows that f(ci) = b i.

… f ni

n(c ) = b1 … bn and hence that f(S

spectrum of f(T) = f 1(T1) … f n(Tn). Infact

1

1 1

1 r {b , ,b }! … n

n n

1 r {b , ,b }! = f 1(S1) … f

t t

t t

m n b bz ; t = 1, 2, …, n; where mt z nt. Let Xm = X

nmX be the set of indices i = (i1, …, in) such that

f(ci) = f 11

1

i(c ) … f nn

n

i(c ) =1m b …

nm b .

LetPm =

1mP … nmP

characteristic n-vectors belonging to the n-charac

bm of f(T) i.e.,1

1

m b … n

n

m b of f 1(T1) …

f(T) = f 1(T1) … f n(Tn)

=1

1 1

1

r

1 1m m

m 1

b P¦ …

n

n n

n

r

n nm m

m 1

b P¦

is the n-spectral resolution of f(T) = f 1(T1) !suppose U = U1 … Un is a n-unitary trans

onto V' and that T' = U TU –1, i.e., Tc1 … T

… UnTn 1nU . Then the equation T = c i.eT h ld d if d l



Tnn = c11 … cnn holds good if and onlyi.e.,

1 1 1 n n nT U T Uc cD D! = c1U11 … cnUnn

… Sn is the n-spectrum of Tc = Tc1 … Tc

… Un maps each n-characteristic n-subspace Tn onto the corresponding n-subspace for T

earlier results we see that

j j

j

T c Ec c ¦

1 nT Tc c ! =1 1

1

1 1

j j

j

c (E )c ¦ ! n j

c¦Here

E' =1 n

1 n

j j(E ) (E )c c ! .

U11 n

1 1 n 1

j 1 n j nE U U E U ! is the n-spectral res

1 nT Tc c ! . Hence

=1 1

1

1 1 1

1 j 1 j 1

j

f (c ) U E U¦ …

n n

n

n n 1

n j n j n

j

f (c )U E U¦

= U1 1 1

1

1 1 11 j j 1

j

f (c )(E ) U¦ … Unn

n

nn j

j

f (c )(E¦= U1f 1(T1)

1

1U ! Unf n(Tn)1

nU

= Uf(T) U-1.

In view of the above theorem we have the following



COROLLARY 2.8: With the assumption of the abo

suppose T = T 1 … T n. is represented in an n-or

B = B1 … Bn =1

1 1

1 n{ }D D! … n

n n

1 n{D D!

diagonal matrix D = D1 … Dn with entries1

1

{d

1{ }n

n n

nd d ! . Then in the n-basis B, f(T) = f 1(T 1 )

is represented by the n-diagonal matrix f(D) = f 1(D

f n(Dn ) with entries1 1

1 1

1 1{ ( ), , ( )}n n f d f d ! … {f

( )}n n

n

n n f d . If B c =1 1

1{( ) , , ( ) }nD D c c! … 1{( )nD c

is any other n-ordered n-basis and P the n-matrix su

t t t t

t

t t t

j i j i

i

P E D ¦

for t = 1, 2, …, n then P -1(f (D)) P is the n-basis Bc .

Proof: For each n-index i = (i1, …, in) there is a uniqu= (j1, …, jn) such that 1 d jt, n p d K t; t = 1

i ij i

i

d P D¦ =1 1 1 1

1

1 1 1

i i j i

i

d P D¦ … n n n

n

n n

i i j

i

d P D¦

= ij i

i

(DP) D¦

=1 1 1

1

1 1i j i

i

(D P ) D¦ … n(D¦= 1

ij Kt K

i K

(DP) P cD¦ ¦

=1 1 1 1 1

1 1

1 1

i j K t K

i K

(DP) P ( ) cD¦ ¦ …

n n n n n

1 n

i j K t K

i K

(DP) P ( ) cD¦ ¦



n ni K ¦ ¦

=1 1 1

1

1 1

1 1 1 K j K

K

(P D P ) ( ) cD¦ …

n n n

n

1 1 n

n n n K j K

K

(P D P ) ( ) cD¦ .

Under the above conditions we have f(A) = f 1(A

= P –1f(D) P = 1

1 1 1 1P f (D )P … 1

n n n nP f (D )P .

The reader is expected to derive other intere

results to usual vector spaces for the n-vector spa Now we proceed onto define the notion of b

for n-vector spaces of type II.

DEFINITION 2.8: Let V = V 1 … V n be a n-ve

the n-field F = F 1 … F n. A bilinear n-for

function f = f 1 … f n which assigns to each

pairs of n vectors in V a n scalar f( ) = f

from V × V into F = F1 … Fn i.e., (V1 × V1) Vn) into F1 … Fn i.e., f: V × V o F is f 1 …

V1) … (Vn × Vn) o F1 … Fn with f t: Vt × V

= 1, 2, …, n which is a n-linear function of e

arguments when the other is fixed. The n-zero funcn-function from V × V into F is clearly a bilinear n

also true that any bilinear combination of bilinear n-

is again a bilinear n-form.

Thus all bilinear n-forms on V is a n-subspac

functions from V × V into F. We shall denote th

bilinear n-forms on V by Ln (V, V, F) = L(V1, V1, L(V V F )



L(Vn, Vn, Fn).

DEFINITION 2.9: Let V = V 1 … V n be a finite

dimensional n-vector space and let B = 1

1{ , ,D D !

1{ , , }n

n n

nD D ! = (B1 … Bn ) be a n-ordered n-ba

= f 1 … f n is a bilinear n-form on V then the n-mthe ordered n-basis B, is (n1 × n1 , …, nn × nn ) n-ma

… An with entries ( , )t t t t

t t t

i j t i j A f D D . At tim

denote the n-matrix by [f] B =11[ ] B f … [ ]

nn B .

THEOREM 2.16: Let V = V 1 … V n be a

dimensional n-vector space over the n-field F = F 1

For each ordered n-basis B = B1 … Bn of V, thwhich associates with each bilinear n-form on V its

the ordered n-basis B = B1 … Bn is an n-isomthe n-space Ln(V, V, F) = L(V 1 , V 1 , F 1 ) … L(V n ,h f ( )

i.e., (c1f 1 + g1)1 1

1 1

i j( , )D D … (cnf n + gn)n

n

i( ,D D

= [c1f 11 1

1 1

i j( , )D D + g11 1

1 1

i j( , )D D ] …

[cnf nn n

n n

i j( , )D D + gnn n

n n

i j( , )D D ].

This simply says [cf + g]B = c[f]B + [g]B i.e.,

1

1

1 1 B[c f g ] … n

n

n n B[c f g ]

=1 1

1

1 B 1 B(c [f ] [g ] ) … n

n(c [f

We leave the following corollary for the reader t



We leave the following corollary for the reader t

COROLLARY 2.9: If B =1

1 1

1{ , , }nD D ! … {D

n-ordered n-basis for V = V 1 … V n and

B* =1

1 11{ , , }n L L! … 1{ , ,n n

n L L!

is the dual n-basis for V * = *

1V … *

nV , t

bilinear n-forms f ij(, ) = Li() L j(); i.e.,

1 1

1

1 1( , ) ( , )n n

n

i j i j n n f f D E D E ! =

1 1

1 1

1( ) ( ) ( ) (

n n

n n

i i j i n j n L L L LD E D E !

1 d it , jt d nt ; t = 1, 2, …, n, forms a n-basis for th

V, F) = L(V 1 , V 1 , F 1 ) … L(V n , V n , F n ) is2

1(n

THEOREM 2.17: Let f = f 1 … f n be a bilinea

finite dimensional n-vector space V of n-dimennn ). Let R f and L f be a n-linear transformationdefined by (L ) = f( ) = (R ) i e

DEFINITION 2.10: If f = f 1 … f n is a bilinear n-f

finite dimensional n-vectors space V and n-rank of

tuple of integers (r 1 , …, r n ); (r 1 , …, r n ) = n-rank L f =

…, rank n

n

f L ) and n-rank R f = (rank 1

1

f R , …, rank

n

f R

We state the following corollaries and the reader is egive the proof.

COROLLARY 2.10: The n-rank of a bilinear n-form i

the n-rank of the n-matrix of the n-form in any ordere



COROLLARY 2.11: If f = f 1 … f n is a bilinear n-

(n1 , …, nn ) dimensional n-vector space V = V 1


a. n-rank f = (n1 , …, nn ) = (rank f 1 , …, rank f n ).b. For each non zero = 1 … n in V th

1 … n in V such that f(, ) z 0 …

1 ) … f n(n , n ) = 0 … 0.

c. For each non zero = 1 … n in V the

1 … n in V such that f(,) = 0

f 1(1 , 1 ) … f n(n , n ) = 0 … 0.

Now we proceed onto define the non degenerate of aform.

DEFINITION 2.11: A bilinear n-form f = f 1 … vector space V = V 1 … V n is called non-deg

If f = f 1 … f n is a n-symmetric bilin

quadratic n-form associated with f is the function

qn from V = V 1 … V n onto F = F 1 …

q() = q1(1 ) … qn(n ) = f(, ) = f 1(1 ,

n ). If V is a real n-vector space an n-inner prod symmetric bilinear form f on V which satisfies f

0 i.e., f 1(1 , 1 ) … f n(n , n ) > (0 …

i z 0 for i = 1, 2, …, t. A bilinear n-form in wh for each i = 1, 2, …, n is called n-positive definit

So two n-vectors , in V are n-orthogonal

a n-inner product f = f 1 f 2 … f n if f(, )

i e f( ) = f1(1 1) f ( ) = 0



i.e., f(, ) = f 1(1 , 1 ) … f n(n , n ) = 0 …The quadratic n-form q() = f(, ) takes on

values.

The following theorem is significant on its own.

THEOREM 2.18: Let V = V 1 … V n be a n-ve

the n-field F = F 1 … F n each F i of charact

1, 2, …, n and let f = f 1 … f n be a n-symm form on V. Then there is an ordered n-basis for

represented by a diagonal n-matrix.

Proof: What we need to find is an ordered n-bas

Bn =1

1 1

1 n{ }D D! … n

n n

1 n{ }D D! such that

f(i, j) = f 11 1

1 1

i j( , )D D … f nn n

n n

i j( , )D D

= 0 … 0for it z jt; t = 1, 2, …, n. If f = f 1 … f n = 0

n n n n n n

1 1q ( ) q ( )

4 4

ª ºD E D E« »¬ ¼.

f = f 1 … f

n

= 0 … 0.

Thus there is n-vector = 1 … n in V such th

q() z 0 … 0 i.e.,

f 1(1,1) … f n(n, n) = q1(1) … q

z 0 … 0.

Let W be the one-dimensional n-subspace of spanned by = 1 … n and let WA be the s



vectors = 1 … n in V such that

f(, ) = f 1 (1, 1) … f n(n, n)

= 0 … 0.

Now we claim W WA = V i.e., 1 1W WA !

V = V1 … Vn. Certainly the n-subspaces W and WA are inde

typical n-vector in V is c where c is a n-scalar. If

i.e.,

c = (c11 … cnn) WA = 1W WA !

thenf 1(c11, c11) … f n(cnn, cnn)

= 2

1c f 1(1, 1) … 2

nc f n(n, n)

= 0 … 0.

But f i(

i,

i) z 0 for every i, i = 1, 2, …, n thus

Also each n-vector in V is the sum of a n-vector in

i A l b

f(, ) = f(, J) – f ( , )

f ( , )f ( , )

J DD D

D D

and since f is n-symmetric, f(, ) = 0 … the n-subspace WA. The expression

f ( , )

f ( , )

J DJ D E

D D

shows that V = W + WA. Then n-restriction of symmetric bilinear n-form on WA

= 1WA !



y 1

has (n1 – 1, …, nn – 1) dimension we may assum

that WA has a n-basis1

1 1 n

2 n 2{ , , } { ,D D D ! ! !

t t

t t

t i jf ( , ) 0D D ; it z jt , it t 2, jt t 2 and t = 1, 2,

= we obtain a n-basis1

1 1 n

1 n 1{ , , } {D D D! !

such that f(i, j) = f 11 1 n n

1 1 n n

i j n i j( , ) f ( , )D D D D!

for it z jt; t = 1, 2, …, n.

THEOREM 2.19: Let V = V 1 … V n bedimensional n-vector space over the n-field of re

let f = f 1 … f n be the n-symmetric bilinewhich has n-rank (r 1 , …, r n ). Then there is an n-

1

1 1

1{ }n E E ! ,2

2 2

1{ }n E E ! … 1{ }n

n n

n E E ! for V

in which the n-matrix of f = f 1 … f n is n-dia

that

f( ) f( ) ( )fE E E E

Now for complex vector spaces we in gene

difficult to define n-vector spaces of type II as it ha

the algebraically closed field. Further when n hap

arbitrarily very large the problem of defining n-vecto

type II is very difficult.First we shall call the field 1 nF F Fc c !

special algebraically closed n-field of the n-field F =

Fn if and only if each iFc happens to be an algebrai

field of Fi for a specific and fixed characteristic poly

Vi over Fi relative to a fixed transformation Ti of Tof every i.



Chapter Three

SUGGESTED PROBLEMS



In this chapter we suggest nearly 120 problems

spaces of type II which will be useful for

understand this concept.

1. Find a 4-basis of the 4-vector space V =

V4 of (3, 4, 2, 5) dimension over the

Q 3 Q 5 Q 7 . Find a 4-s

2. Given a 5-vector space V = V1 V2 over Q 2 Q 3, 5 Q 7

3. Let V = V1 V2 V3 V4 be a 4-vector spa

4-field (Q 2 Q 7 Z11 Z2) of di

4, 5, 6).

a. Find a 4-subset of V which is a depende

of V.

b. Give an illustration of a 4-subset of Vindependent 4-subset of V but is not a 4-

c. Give a 4-subset of V which is a 4-basis o

d. Give a 4-subset of V which is semi n-d

V.



e. Find a 4-subspace of V of dimension

over the 4-field F where F = (Q 2

Z11 Z2) is a field.

4. Given V = V1 V2 V3 is a 3-vector spac

over the 3-field F = Q Z2 Z5 of dimension

Suppose

A = A1 A2 A3

=

3 1

2 0 3 00 1 6 0 11 2 0 1

3 1 0 1 00 0 1 3

1 0 1 0 11 1 0 0

1 0

ª

ª º « ª º « » « « » « » « « » « » « « » « »¬ ¼ « ¬ ¼ « ¬

is the 3-matrix find the 3-linear operator relate

Find a 4-linear transformation T from V

will give way to a nontrivial 4-null subspa

6. Let V = V1 V2 V3 V4 V5 be a

over the 5-field F = Q 3 Q 2 Z2 of type II of dimension (3, 4, 2, 5, 6).

a. Find a 5-linear operator T on V su

invertible.

b. Find a 5-linear operator T on V suc

5-invertible.

c. Prove 5-rank T + 5 nullity T = (3, 4



T.

d. Find a T which is onto on V and fin

T.

7. Let V = V1 V2 V3 V4 be a (5, 4

space over the 4-field Z2 Z7 Q 3

V*. Obtain for any nontrivial 4-basis B its

8. Given V* is a (3, 4, 5, 7) dimensional dua

4-field F = Z2 Z7 Z3 Z5. Find V. Wh

9. Let V = V1 V2 V3 V4 be a 4-vecto

5, 6) dimension over the 4-field F = Z5 Suppose W is a 4-subspace of (2, 3, 4, 5)

the 4-field F. Prove dim W + dim Wo

explicitly Wo. What is Woo? Is Wo a 4-suV*? Justify your claim

11. Given V = V1 V2 V3 V4 V5 is a 5-sp

2, 3, 7) dimension over the 5-field F = Q 3

Q

2Z7. Find a 5-hypersubspace of V

Prove W is a 5-hypersubspace of V. What is

5-basis for V and its dual 5-basis. Find a 5-b

What is its dual 5-basis?

12. Let V = V1 V2 V3 V4 be a 4-space ove

F = F1 F2 F3 F4 of dimension (7, 6, 5, 4)Let W1 and W2 be any two 4 subspaces of (4,

(3 5 4 2) dimensions respectively of V Fin



(3, 5, 4, 2) dimensions respectively of V. Fino

2W . Is o

1W = o

2W ? Find a 4-basis of W1 and W

dual 4-basis. Is W1 a 4-hypersubspace of V?

claim! Find a 4-hyper subspace of V.

13. Let V = V1 V2 V3 V4 V5 be a (2,

dimensional 5-vector space over the 5-field F

Z5 Z7 Q. Find V*. Find a 5-basis and

basis. Define a 5-isomorphism from V into V

W1 W2 … W5 is a (1, 2, 3, 4, 5), 5-subfind the n-annihilator space of V. Is W a 5-hyp

V? Can V have any other 5-hyper space other t

14. Let V = V1 V2 V3 V4 be a 4-space of (

dimension over the 4-field F = Z3 Z5 Z7

4-transformation T on V such that rank (Tt

)(Assume T is a 4-linear transformation which

17. Given A = A1 A2 A3 where A1 is a

matrices with entries from Q 3 . A

algebra over Q 3 . A2 = {All poly

variable x with coefficient from Z7}; A2 isover Z7 and A3 = {set of all 5 u 5 matri

from Z2}. A3 a linear algebra over Z2.

algebra over the 3-field F = Q 3 uZ7

commutative, 3-linear algebra over F? Doe

3-identity?

18. Define a n-sublinear algebra of a n-linear a

n field F



n-field F.

19. Give an example of a 4-sublinear algebra

algebra over the 4-field.

20. Give an example of an n-vector space of

not an n-linear algebra of type II.

21. Give an example of an n-commutative n

over an n-field F (take n = 7).

22. Give an example of a 5-linear algebra w

commutative over the 5-field F.

23. Let A = A1 A2 A3 where A1 is a s

matrices over Q, A2 = all polynomials in

with coefficients from Z2 andn- ½

25. Using the 5-field F = Z2 =7 =3 Z5 =the 5-vector space = V = Z2 [x] Z7 [x] Z3

Z11 [x]. Verify

(i) (cf + g)D = cf (x) + g (D(ii) (fg)D = f(D) g(D)

For C = C1 C2 C3 C4 C5

= 1 5 2 3 10;

f = (x2 + x + 1) (3x3 + 5x + 1) (2x4 + x +

2) (10x2 + 5) where x2 + x + 1 Z2[x], 3x3

Z7[x], 2x4 + x + 1 Z3[x], 4x + 2 Z5[x] and

Z11[x] and g(x) = x3

+ 1 x2+ 1 3x

3+ x

2+

2 3



+ x + 1 7x2

+ 4x + 5 where x3

+ 1 Z2[x

Z7[x], 3x3

+ x2

+ x + 1 Z3[x], 4x5

+ x + 1

7x2

+ 4x + 5 Z11[x].

26. Let V = Z2[x] Z7[x] Q[x] be a 3-linear a

the 3-field F = Z2 Z7 Q. Let M = M1 0the 3-ideal generated by {¢x2 + 1, x5 + 2x + 1

+ x + 1, x + 5, 7x3 + 1²} {¢x2 + 1, 7x3 + 5x

Is M a 3-principal ideal of V.

27. Prove in the 5-linear algebra of 5-polynomial

Z2[x] Q[x] Z7[x] Z17[x] over the 5-fi

Q Z7 =17 every 5-polynomial p = p1 pcan be made monic. Find a nontrivial 5-ideal o

28. Let A = A1 A2 … A6 = Z3[x] Z7[x]Q[x] Z [x] Z [x] be a 6 linear algebra

= 1, 2, …, n. An ideal M = M1 … Mn

semi maximal if and only if there exists at

of ideals in M, m d n which are maxim

them are minimal. Similarly we say M = M

n-semi minimal if and only if M containsideals which are minimal, p d n and none

M are maximal.

29. Give an example of an n-maximal ideal.

30. Give an example of an n-semi maximal idn-maximal.



31. Give an example of an n-semi minimal id

an n-minimal ideal.

32. Let A = Z3[x] Z2[x] Z7[x] Z5[xalgebra over the 4-field F = Z3 Z2 Zexample of a 4-maximal ideal of A. Ca

minimal ideal? Justify your claim. Does A

maximal ideal? Can A have a 4-semi minim

33. Give an example of a n-linear algebra whsemi maximal and n-semi minimal ideals.

34. Give an example of a n-linear algebra wh

which is not a n-principal ideal? Is this po

… An where each Ai is Fi[x] where A

the n-field F = F1 … Fn; for i = 1, 2, …

36. Find the 4-characteristic polynomial and

polynomial for any 4-linear operator T = T1 T4 defined on the 4-vector space V = V1 V2

where V is a (3, 4, 2, 5) dimensional over the

Z5 Q Z7 Z2.a. Find a T on V so that the 4-characteristic

is the same as 4-minmal polynomial.

b. Define a T on the V so that T is

diagonializable operator on V.

c. For a T in which the 4-chracteristic po

different from 4-minimal polynomial findof polynomials over the 4-field F which

T.



d. For every T can T be 4-diagonalizable.

claim.

37. Let

A =

1 0 04 0 1 1

0 1 0 0 1 00 1 1 3

1 0 1 1 0 11 0 0 0

0 0 2 0 0 1

2 1 0 0 1 1 0

ª ª º « ª º « » « « » « » « « » « » « « » « »¬ ¼ « ¬ ¼ « ¬

0 1 0 2 3 5

7 0 0 1 0 0

1 0 1 0 2 0 0 1

2 6 0 0 1 1 0 0

ª º« »« »« »ª º

« »« »¬ ¼ « »

characteristic space of all D such that TDto A.

38. Let A = A1 … A5 be a (5 u 5, 3 u3, 2

6), 5-matrix over the 5-field F = Z2 Z3 If A

2= A i.e., Ai

2= Ai for i = 1, 2, …, 5

similar to a 5-diagonal matrix. What can

the 5-characteristic vectors and 5-chara

associated with A.

39. Let V = V1 … V6 be a 6-vector spacZ2 Q Z7 Z5 Z3 Z11. Let T be a 6

operator on V. Let W = W1 … W6



operator on V. Let W W1 … W6

subspace of V under T. Prove the 6-rest

TW is 6-diagonalizable.

40. Let V = V1 V2 V3 be a 3-vector spfield, F = Q Z2 Z5 of (5, 3, 4) dimen

W = W1 W2 W3 be a 3-invariant 3

( Hint : Find T and find its 3-subspace W wunder T). Prove that the 3-minimal 3-pol

3-restriction 3-operator TW divides

polynomial for T. Do this without referring

41. Let V = V1 … Vn be a (3 u 3, 2 u,

space of matrices over the 4-field F = Z2

Let T and U be 4-linear operators on V def

T (B) = AB

i.e., T(B1 B2 B3 B4) = A1B1 …



1

1 n

n1 1 n n

1 1 n n 1 n i

j 1

f x x f x ...x c x

¦ ! !

b. Show that the n-dual space W* of W

identified with n-linear functionals.

1

1 1 n

1 1 n n 1f x x f x x ! ! !

=1 1

1 1 1 1 n n

1 1 n n 1 1c x ... c x ... c x ... c

on 1 2 nn n n

1 2 nF F ... F which satisfy

for i = 1, 2, …, n.

57. Let W = W1 … Wn be a n-subspace



…, nn) dimensional n-vector space over V

Vn and if

^ ` ̂ ` ^1 2

1 1 2 2 n

1 r 1 r 1g g g g g ! ! !

is a basis for Wo = o o

1 nW ... W then

W =1 n

i i11 n

r r1 n

g g gi i 1 i 1

N N N

!

where ^ ` ^ `1 n

1 1 n n

1 r 1 rN N N N ! ! ! is the

space of the n-linear functionals

f = f 1 … f n = ^ ` ^1

1 1

1 rf f ! !

and

^ ` ^ `1 n

1 1 n n

1 r 1 rg g g g ! ! !

is the n-linear combination of the n-linear

… f n.

cf(x) = c1f 1(x) … cnf n(x)where c = c1 = F1 … Fn . Let W = W1 … Wn be

nn) dimensional space of V(S; F) = V1(S1; F

Vn(Sn, Fn).

Prove that there exists (n1, …, nn) points

1 2

1 1 2 2 n

1 n 1 n 1x x x x x ! ! ! !

in S = S1 … Sn and (n1, …, nn) functions

1 2 n

1 1 2 2 n n

1 n 1 n 1 nf f f f f f ! ! ! !

in W such that

t t t

i j ij

f x G ; i = 1, 2, …, n.

59. Let V = V1 … Vn be a n-vector space ove

F F F d l T T T b



F = F1 … Fn and let T = T1 … Tn b

operator on V. Let C = C1 … Cn be

suppose there is a non zero n-vector D = D1 V such that TD = CD i.e., T1D1 = C1D1, …, TnDProve that there is a non-zero n-linear functio

… f n on V such that Ttf = Cf i.e., t

1 1T f ...

… cnf n.

60. Let A = A1 … An be a (m1 u m1, …, mmixed rectangular matrix over the n-field F =

Fn with real entries. Prove that A = 0 …Ai = (0) for i = 1, 2, …, n) if and only if the

(0) i.e., A1tA1 … An

tAn = 0 … 0.

61. Let (n1, …, nn) be n-tuple of positive integers athe n space of all n polynomials functions ove

62. Let V = V1 … Vn be a n-vector space

F = F1 … Fn. Show that for a n-linear

T = T1 … Tn on T oTtis an n-i

Ln

(V, V) onto Ln

(V*, V*);i.e., T = T1 … Tn oTt = t

1T .

isomorphism of Ln(V, V) = L(V1, V1) onto L

n(V

*, V

*) = * * *

1 1 nL V ,V L V , !

63. Let V = V1 … Vn be a n-vector spaceF = F1 … Fn, where V is a n-space

un, …, nn unn); n-matrices with entries



F = F1 … Fn.

a. If B = B1 … Bn is a fixed (n1 u n1

unn) n-matrix define a n-function f B = f 1on V by f B(A) = trace(B

tA);

f B(A) =1 nB 1 Bf (A ) f (A !

= trace t

1 1B A trace B !

where A = A1 … An V = V1 f B is a n-linear functional on V.

b. Show that every n-linear functional on

form i.e., is f B for some B.

c. Show that B o1 n1B nBf f ! i.e., B1

1 n1B nBf f ! i.e., Bi oiiBf ; i = 1, 2,

isomorphism of V onto V*.

L1(f 1) L

1(g1) … L

n(f n) L

n(gn).

Prove that if L = L1 … Ln is any n-linea

on F[x] = F1[x] … Fn[x] such that L(fg)

for all f, g then L = 0 … (0) or there is a tn such that L(f) = f(t) for all f. i.e., L1(f 1) = f 1(t) … f n(t).

65. Let K = K1 … Kn be the n-subfield of a n-

… Fn . Suppose f = f 1 … f n and g =

gn be n-polynomials in K[x] = K1[x] … MK =

1 nK KM M ! be the n-ideal generated

in K[x] Let MF = F FM M ! be the



in K[x]. Let MF1 nF FM M ! be the

generated in F[x] = F1[x] … Fn[x]. Show

MF have the same n-monic generator. Suppos

polynomial in F[x] = F1[x] … Fn[x], if Mideal of F[x] i.e., MF =

1 nF FM M ! . Find

under which the n-ideal Mk of K[x] = K1[x] can be formed.

66. Let F = F1 … Fn be a n-field. F[x] = F1

Fn[x]. Show that the intersection of any nu

ideals in F[x] is again a n-ideal.

67. Prove the following generalization of the Tayl

for n-polynomials. Let f, g, h be n-polynomi

n-subfield of complex numbers with n-deg f =deg fn) d (n1 nn) where f = f1 fn

69. Let V = V1 … Vn be a (n1, n2, …, nn)

vector space. Let W1 = 1

1W W !

subspace of V. Prove that there exists a n

1 n

2 2W W ! of V such that V = W1

V = W1 W2

= 1 1 2 2 n

1 2 1 2 1W W W W W !

70. Let V = V1 … Vn be a (n1, …, nn) finn-vector space and let W1, …, Wn be n-

such that V = W1 + … + Wn i.e., V = V

1 1 2 2W W W W



1 21 k 1 kW W W W ! ! !

and n dim V = (n dim W1 + … + n dim W

= (dim

1

1W + … + dim 1

1

k W , dim

2

1W + …dim n

1W + … + dimn

n

k W ). Prove that V = W

1 2

1 1 2 2

1 k 1 kW W W W ! ! !

71. If E1 and E2 are n-projections E1 =

1

1

E !

= 1 n

2 2E E ! on independent n-subsp

+ E2 an n-projection? Justify your claim.

72. If E1 , …, En are n-projectors of a n-vecto

that E1

+ … + En

= I.

i.e., 1 nk k 1 1

1 1E E E E ! ! !

74. Obtain some interesting properties about n-ve

of type II which are not true in case of n-vecto

type I.

75. Let T be a n-linear operator on the n-vector type II which n-commutes with every n

operator on V. What can you say about T?

76. If N = N1 … Nn is a n-nilpotent linear op

(n1, …, nn) dimensional n-vector space V = V1

then the n-characteristic polynomial for N is (1n

x ).

77 Let V = V V be a (n n n ) dim



77. Let V = V1 … Vn be a (n1, n2, …, nn) dim

vector space over the n-field F = F1 … Fn

… Tn be a n-linear operator on V such th

= (rank T1, …, rank Tn) = (1, 1, …, 1). Prove tis n-diagonalizable or T is n-nilpotent,

simultaneously.

78. Let V = V1 … Vn be a (n1, n2, …, nn) dim

vector space over the n-field F = F1 … F

T1 … Tn be a n-operator on V. Suppose T

with every n-diagonalizable operator on V, i

commutes with every ni-diagonalizable operat

i = 1, 2, …, n then prove T is a n-scalar multip

identity operator on V.

79. Let T be a n-linear operator on the (n1,

t

jW is the null space of jr

tj tp (T ) , j = 1, 2,

2, …, n. Let Wr

= r r

1 nW W ! be any n

which is n-invariant under T. Prove that

Wr

= r r

1 1 2 2W W W W ...

80. Define some new properties on the n-v

type II relating to the n-linear operators.

81. Compare the n-linear operators on n-vec

type II and n-vector space of type I. Istransformation of type I always be a n-line

type I n-vector space?



82. State and prove Bessel’s inequality in c

spaces.

83. Derive Gram-Schmidt orthogonalization

vector spaces of type II.

84. Define for a 3-vector space V = (V1 V2

5) dimension over the 3-field F = Q

distinct 3-inner products.

85. Let V = V1 V2 V3 be a 3-space

dimension over the 3-field F = Q 2

Q 7 . Find L3(V, V, F) = L(V1, V1, Q

V2, Q 3 )L (V3, V3, Q 7 ).

special 5-subfield F = F1 F2 F3 F4 complex field (such that Fi zF j if i z j, 1 d i, j da. V is of (7, 3, 4, 5, 6) dimension over F =

F5.

b. Prove P2

= P is a 5-projection.c. Find 5-rank P and 5-nullily P. Is 5 rank P

10, 15) and 5-nullity P = (28, 6, 10, 15, 21)

88. Let V = V1 … V4 be a finite (4, 5, 3, 6

vector space over the 4-field Q 2 Q 3

Q 7 and f a symmetric bilinear 4-form

each 4-subspace W of V. Let WA

be the set of

D = D D D D in V such that f(DE) =



( E)

0 for E in W. Show that

a. WA

is a 4-subspace.

b. V = {0}A {0}A {0}A {0}Ac. V

A= {0} {0} {0} {0} if and only

degenerate! Can this occur?

d. 4-rank f = 4-dim V – 4-dim VA

.

e. If 4-dim V = (n1, n2, n3, n4) and 4-dim W

m3, m4) then 4-dim WA t (n1 – m1, n2 – m2

–m4).f. Can the 4-restriction of f to W be a non-d

W WA= {0} {0} {0} {0}?

89. Prove if U and T any two normal n opera

commute on a n-vector space over a n-field

prove U + T and UT are also normal n-operato

92. Prove that every positive n-matrix is th

positive n-matrix.

93. Prove that a normal and nilpotent n-operatoperator.

94. If T = T1 … Tn is a normal n-operato

n-characteristic n-vectors for T which are

distinct n-characteristic values are n-orthog

95. Let V = V1 … Vn be a (n1, n2, …, nn)

inner product space over a n-field F = F1

each n-vector D, E in V let TDE =1TD



, E DE D

(where T = T1 … T

n, D = D1 …

… En) be a linear n-operator on V defin

(J / ED i.e., TDE (J) =1 1

1

, 1T ( ) TD E J !

/ ED … (Jn / EnDn. Show that

a. *

, ,T TD E E D .

b. Trace (TDE) = (D / E).

c. (TDE) (TJG) = TD(E / J)Gd. Under what conditions is TDE n-self ad

96. Let V = V1 … Vn be a (n1, n2, …, nn)

inner product space over the n-field F = F

let Ln(V, V) = L(V1, V1) … L(Vn, Vn

of linear n-operators on V. Show that therinner product on Ln (V V) with the proper

tr(AnBn*) where A = A1 $n and B = B

are (n1 u n1, …, nn unn) n-matrix.

97. Let V = V1 … Vn be a n-inner product sp

E = E1 … En be an idempotent linear n-V. i.e., E2 = E prove E is n-self adjoint if and

= E*E i.e., E1 E1* … En En

* = E1*E1 …

98. Show that the product of two self n-adjoint

self n-adjoint if and only if the two operators c

99. Let V = V1 … Vn be a finite (n1,

dimensional n-inner product vector space of

Let T = T1 … Tn be a n-linear operator o



1 n p

that the n-range of T* is the n-orthogonal com

the n-nullspace of T.

100. Let V be a finite (n1, n2, …, nn) dimensional in

space over the n-field F and T a n-linear opera

T is n-invertible show that T* is n-invertible a

(T-1)*.

101. Let V = V1 … Vn be a n-inner product spn-field F = F1 … Fn. Eand J be fixed n-v

Show that TD= (D / E)J defines a n-linear ope

Show that T has an n-adjoint and describe T* e

102. Let V = V1 … Vn be an n-inner product

polynomials of degree less than or equal to

For D = D1 … Dn and E E1 … Ei.e., ||DE||

2 + ||D – E||2 … ||DnEn

= 2||D1||2

+ 2||E||2 … 2||Dn||

2+ 2|

104. Let V = V1 … Vn be a n-vector spaceF = F1 … Fn. Show that the sum

product on V is an n-inner product on V. I

of two n-inner products an n-inner produc

positive multiple of an n-inner product

product.

105. Derive n-polarization identity for a n-vect

… Vn over the n-field F = F1 …

standard n-inner product on V.



106. Let A = A1 … An be a (n1 u n1, …, nn

with entries from n-field F = F1 …

^ ` ^ `1 n

1 1 n n

1 n 1 nf f f f ! ! ! be the n-dia

the n-normal form of xI – A = xI1 – A1 For which n-matrix A is 1 n

1 1f , , f ! z (1,

107. Let T = T1 … Tn be a linear n-oper

(n1, …, nn) dimensional vector space over

F1 … Fn and A = A1 … An

associated with T in some ordered n-basi

n-cyclic vector if and only if the n-d

(n1 – 1)u (n1 – 1), …, (nn – 1) unn – 1) nxI – A are relatively prime.

complementary T-n-invariant n-subspace

if R is n-independent of the n-null space N

Nn of T.

b. Prove if R and N are n-independent N is th

n-variant n-subspace complementary to R.

110. Let T = T1 … Tn be a n-linear operato

space V = V1 … Vn. If f = f 1 … polynomial over the n-field F = F1 … Fn a

Dn = D and let f(D) = f(T)D

i.e., f 1(D1) … f n(Dn)= f 1(T1)D1 f n(Tn)Dn.

If ^ ` ^ `1 n

1 1 n n

1 k 1 k V V , , V V! ! ! are T-n-invari

1 1



space of V= 1

1 1 n

1 k 1V V V ! ! !

that fV = 1

1 1 n

1 1 1 k n 1f V f V f V ! ! !

111. Let T, V and F are as in the above prob

Suppose D1 … Dn and E E1 … En a

in V which have the same T n-annihilator. Pr

any n-polynomial f = f 1 … f n the n-vecto

… f nDn and f E = f 1E1 … f nEn have annihilator.

112. If T = T1 … Tn is a n- diagonalizable oper

vector space then every T n-invariant n-subsp

complementary T-n-invariant subspace.

… Tn be a n-linear operator on V. Wh

zero n-vector in V a n-cyclic vector for T?

is the case if and only if the n-characterist

for T is n-irreducible over F.

115. Let T = T1 … Tn be a n-linear operato

…, nn) dimensional n-vector space over th

II. Prove that there exists a n-vector D = (

in V with this property. If f is a n-polynom

… f n and f(T)D = 0 … 0 i.e., f 1

f n(Tn)Dn = 0 … 0 then f(T) = f 1(T1) … 0. (such a n-vector is called a sep

for the algebra of n-polynomials in T). W

cyclic vector give a direct proof that any n



is a separating n-vector for the algebra of

in T.

116. Let T = T1 … Tn be a n-linear ope

vector space V = V1 … Vn of

dimension over the n-field F = F1 … F

a. The n-minimal polynomial for T is

irreducible n-polynomial.

b. The minimal n-polynomial is characteristic n-polynomial. Then

nontrivial T-n-invariant n-subspace

complementary T-n-invariant n-subspa

117. Let A = A1 … An be a (n1 u n1, …, n

with real entries such that A2 + I = 21A I

118. Let A = A1 … An be a (m1 u n1, …,

matrix over the n-field F = F1 … Fn and c

n-system of n-equation AX = Y i.e., A1X1 …

Y1 … Yn. Prove that this n-system of equ

n-solution if and only if the row n-rank of Athe row n-rank of the augmented n-matrix

system.

119. If Pn = nnn

1 nP P ! is a stochastic n-mat

stochastic n-matrix? Show if A = A1 …

stochastic n-matrix then (1, …, 1) is an n-eigA.

120. Derive Chapman Kolmogorov equation for



1 n

1 1 n n

(n ) (n )n

ij i j i jp p p !

=1 n

1 1 1 1 n n n n

1 1 n n

(n 1) (n 1)

i j k j i j k jk s k s

p p p p

¦ ¦"

S = S1 … Sn is an associated n-set.

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7. BURROW, M., Representation Theory of

11. GREUB, W.H., Linear Algebra, Fourth Edition

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14. HARVEY E. ROSE, Linear Algebra, Bir Khau

2002.

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16. HERSTEIN, I.N., and DAVID J. WINTER, Matrix Linear Algebra, Maxwell Pub., 1989.



g , ,

17. HERSTEIN, I.N., Topics in Algebra, John Wiley,

18. HOFFMAN, K. and KUNZE, R., Linear algebra, P

of India, 1991.

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25. KOSTRIKIN, A.I, and MANIN, Y. I., Line

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26. LANG, S., Algebra, Addison Wesley, 1967.

27. LAY, D. C., Linear Algebra and its ApplicWesley, 2003.

28. MAC WILLIAM, F.J., and SLOANE N.J.A.,

Error Correcting Codes, North Holland Pub

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30. PLESS, V.S., and HUFFMAN, W. C., HandTheory, Elsevier Science B.V, 1998.



y

31. ROMAN, S., Advanced Linear Algebra, S

New York, 1992.

32. RORRES, C., and ANTON H., Applica Algebra, John Wiley & Sons, 1977.

33. SEMMES, Stephen, Some topics pertaining

linear operators, Novembehttp://arxiv.org/pdf/math.CA/0211171

34. SHANNON, C.E., A MathematicCommunication, Bell Systems Technical Jo

423 and 623-656, 1948.

35. SHILOV, G.E., An Introduction to the Th

38. VASANTHA KANDASAMY and RAJKUMAR, R.

approximations in algebraic bicoding theory,

Journal of Mathematical Sciences, 6, 509-516, 2

39. VASANTHA KANDASAMY and THIRUVEGA

Application of pseudo best approximation to codUltra Sci., 17, 139-144, 2005.

40. VASANTHA KANDASAMY, W.B., Bialgebraic str

Smarandache bialgebraic structures, America

Press, Rehoboth, 2003.

41. VASANTHA KANDASAMY, W.B., Bivector spac

Phy. Sci., 11, 186-190 1999.

42. VASANTHA KANDASAMY, W.B., Linear A



, ,

Smarandache Linear Algebra, Bookman Publish

43. VASANTHA KANDASAMY, W.B., SMARANDACH

and K. ILANTHENRAL, Introduction to bimatrPhoenix, 2005.


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INDEX

B

Bilinear n-form, 187



C

Cayley Hamilton theorem for n-linear operator o

space of

Cyclic n-decomposition, 136-7

F

Finite n- dimensional n-vector space, 13

G

Generalized Cayley Hamilton theorem for n-vect

M

m-subfield, 9-10

N

n-adjoint, 173, 177

n-algebraically closed n-field relative to a n-polynom

n-algebraically closed n-field, 93-4

n-annihilator of a n-subset of a n-vector space of type

n-best approximation, 166-7

n-characteristic value of a n-linear operator of a n-ve

of type II, 8

n-commutative n-linear algebra of type II, 66-7

n-companion n-matrix of the n-polynomial, 134-5



n-conjugate transpose of a n-matrix of T, 173

n-cyclic decomposition, 136-7

n-cyclic n-decomposition , 136-7

n-cyclic n-subspace, 130

n-cyclic n-vector for T, 152

n-diagonalizable n-linear operator, 93

n-diagonalizable normal n-operator, 182

n-divisible, 77-8

n-dual of a n-vector space of type II, 49, 57

n-equation, 182

n-field of characteristic zero, 8

n-field of finite characteristic, 8-9

n-field of mixed characteristic, 8-9

n-field, 7

n-hypersubspace of a n-vector space of type II, 52, 54

n-linear algebra with identity of type II, 66-7

n-linear functional on a n-vector space of type II

n-linear n-combination of n-vectors, 14-5

n-linear operator on a n-vector space of type II, 3

n-linear transformation of type II n-vector spaces

n-linearly dependent n-subset in n-vector space o

n-linearly dependent, 11-2

n-linearly independent n-subset of type II n-vecto

n-mapping, 168

n-monic n-polynomial, 68

n-normality, 183

n-nullity of T, 28

non degenerate, 190

n-ordered pair, 161-2

n-orthogonal projection of a n-subspace, 167-8

h l 163 178 190 1



n-orthogonal, 163, 178, 190-1

n-orthonormal, 163-4

n-positive definite, 190-1

n-range of T, 28

n-rank of T, 28

n-rational form, 136-7

n-reducible n-polynomial over a n-field, 86

n-root or n-zero of a n-polynomial, 68

n-semiprime field, 9-10

n-similar, 47-8

n-spectrum, 182

n-standard inner product, 161-2

n-subfield, 9

n-subspace spanned by n-subspace, 15

n-symmetric bilinear n-form, 190-1

Q

Quadratic n-form, 190-1

Quasi m-subfield, 9-10

S

Semi n-linearly dependent n-subset, 11-2

Semi n-linearly independent n-subset, 11-2

Semi prime n-field, 9-10

Spectral n-resolution, 182

T

T ihil t f t 131 139 140



T-n-annihilator of a n-vector, 131, 139-140

T-n-conductor, 110

ABOUT THE AUTHORS

Dr.W.B.Vasantha Kandasamy is an Associate

Department of Mathematics, Indian Institute Madras, Chennai. In the past decade she has gscholars in the different fields of non-assoc

algebraic coding theory, transportation theory, fuapplications of fuzzy theory of the problems faindustries and cement industries.

She has to her credit 646 research papers.

over 68 M.Sc. and M.Tech. projects. She collaboration projects with the Indian S

Organization and with the Tamil Nadu State AIDS This is her 39th book.

On India's 60th Independence Day, Drconferred the Kalpana Chawla Award for Cour



pEnterprise by the State Government of Tamil Nad

of her sustained fight for social justice in the InTechnology (IIT) Madras and for her contribution (The award, instituted in the memory of Iastronaut Kalpana Chawla who died aboard Columbia). The award carried a cash prize of five highest prize-money for any Indian award) and a She can be contacted at vasanthakandasamy@gm

You can visit her on the web at: http://mat.iitm.ac

Dr. Florentin Smarandache is a Professor of MChair of Math & Sciences Department at the Un

Mexico in USA. He published over 75 books and notes in mathematics, physics, philosophy, psy

literature.I th ti hi h i i b

n-linear algebra of type ii, by w. b. vasantha kandasamy, florentin smarandache

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