1 mac 2103 module 10 lnner product spaces i. 2 rev.f09 learning objectives upon completing this...

1

MAC 2103

Module 10

lnner Product Spaces I

2Rev.F09

Learning ObjectivesUpon completing this module, you should be able to:

1. Define and find the inner product, norm and distance in a given inner product space.

2. Find the cosine of the angle between two vectors and determine if the vectors are orthogonal in an inner product space.

3. Find the orthogonal complement of a subspace of an inner product space.

4. Find a basis for the orthogonal complement of a subspace of ⁿ spanned by a set of row vectors. ℜ

5. Identify the four fundamental matrix spaces of an m x n matrix A, and know the column rank of A, the row rank of A, and the rank of A.

6. Know the equivalent statements of an n x n matrix A.

http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

http://faculty.valenciacc.edu/ashaw/

3Rev.09

General Vector Spaces II


Inner Product Spaces, Inner Products, Inner Product Spaces, Inner Products,

Norm, Distance, Fundamental Matrix Spaces, Norm, Distance, Fundamental Matrix Spaces, Rank, and OrthogonalityRank, and Orthogonality

The major topics in this module:


4Rev.F09

Definition of an Inner Product and Orthogonal Vectors


A. Inner Product: Let

be any function from into that satisfies the following conditions for any u, v, z in V:

and iff

Then defines an inner product for all u, v in V.

u and v in V are Orthogonal Vectors iff

⟨⋅,⋅⟩:V ×V → °

V ×V °

c) ⟨kru,

rv⟩ =k⟨

ru,

rv⟩,

a) ⟨ru,

rv⟩ =⟨

rv,

ru⟩,

d) ⟨rv,

rv⟩ ≥0, ⟨

rv,

rv⟩= 0

rv =0.

⟨ru,

rv⟩

⟨ru,

rv⟩= 0.


5Rev.F09

Definition of a Norm of a Vector


B. Norm: Let

be any function from into that satisfies the following conditions for any u, v in V:

Then defines a norm for all u in V.

The special norm that is induced by an inner product is

⋅ :V → ° + = [0,∞)

V °+

ru

a)ru ≥0, and

ru =0 iff

ru=

r0,

b) sru = s

ru , and

c)ru+

rv ≤

ru +

rv Triangle inequality.

ru = ⟨

ru,

ru⟩.

6Rev.F09

Definition of the Distance Between Two Vectors


C. Distance: Let

be any function from into that satisfies the following conditions for any u, v, w in V:

Then defines the distance for all u, v in V.

The special distance that is induced by a norm is

d(⋅,⋅) :V ×V → ° + =[0,∞)V ×V °

+

d(ru,

rv)

d(ru,

rv)=

ru−

rv = ⟨

ru−

rv,

ru−

rv⟩.

a) d(ru,

rv)≥0, and d(

ru,

rv) =0 iff

ru=

rv,

b) d(ru,

rv) =d(

rv,

ru), and

c) d(ru,

rv) ≤d(

ru,

rw) +d(

rw,

rv) Triangle inequality.

7Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors in ⁿℜ ?


For u and v in ⁿℜ , we define

and

Note: These are our usual norm and distance in ⁿℜ . ⁿℜ is an inner product space with the dot product as its inner product.

ru =⟨

ru,

ru⟩

12 = (

ru⋅

ru) = u1

2 +u22 + ...+un

2 ,

d(ru,

rv)=

ru−

rv

=⟨ru−

rv,

ru−

rv⟩

12 = (

ru−

rv)⋅(

ru−

rv)

= (u1 −v1)2 + (u2 −v2 )

2 + ... + (un −vn)2 .


8Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors in ⁿℜ ? (Cont.)


Example 1: Let u = (1,2,-1) and v = (2,0,1) in ³. Find the ℜinner product, norms, and distance.

⟨ru,

rv⟩=

ru ⋅

rv = u1v1 + u2v2 + u3v3 = (1)(2) + (2)(0) + (−1)(1) = 1,

ru = ⟨

ru,

ru⟩

12 = u1

2 + u22 + u3

2 = 12 + 22 + (−1)2 = 6,rv = ⟨

rv,

rv⟩

12 = v1

2 + v22 + v3

2 = 22 + 02 +12 = 5, and

d(ru,

rv) =

ru −

rv = ⟨

ru −

rv,

ru −

rv⟩

12

= (1− 2)2 + (2 − 0)2 + (−1−1)2 = 9 = 3.

9Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Matrices) in M22?


In M22, the vectors are 2 x 2 matrices. Thus, given two matrices U and V in M22, we define

U =u1 u2

u3 u4

⎡

⎣⎢⎢

⎤

⎦⎥⎥,V =

v1 v2v3 v4

⎡

⎣⎢⎢

⎤

⎦⎥⎥.

Recall: tr(A) = sum of the entries on the main diagonal of A.

⟨U,V⟩= tr(UTV ) = tr(V TU ) = ⟨V ,U⟩= u1v1 + u2v2 + u3v3 + u4v4 .

U =⟨U,U⟩12 = u12 +u2

2 +u32 +u4

2 , and

d(U,V) = U −V =⟨U −V,U −V⟩12 = ⟨U −V,U −V⟩

= (u1 −v1)2 + (u2 −v2 )

2 + (u3 −v3)2 + (u4 −v4 )

2 .

10

Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Matrices) in M22? (Cont.)


Example 2: Let

be matrices in the vector space M22. Find the inner product, norms and distance for U and V in M22.

U =u1 u2

u3 u4

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 2 −1

3 1⎡

⎣⎢

⎤

⎦⎥,V =

v1 v2v3 v4

⎡

⎣⎢⎢

⎤

⎦⎥⎥= 0 9

4 6⎡

⎣⎢

⎤

⎦⎥

⟨U,V⟩= tr(UTV ) = tr(V TU ) = u1v1 + u2v2 + u3v3 + u4v4

= (2)(0) + (3)(4) + (−1)(9) + (1)(6) = 9.Recall: tr(A) = sum of the entries on the main diagonal of A.

11

Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Matrices) in M22? (Cont.)


U =⟨U,U⟩12 = u12 +u2

2 +u32 +u4

2 = 22 + (−1)2 + 32 +12 = 15,

V =⟨V,V⟩12 = u12 +u2

2 +u32 +u4

2 = 02 + 92 + 42 + 62 = 133, and

d(U,V) = U −V =⟨U −V,U −V⟩12 = ⟨U −V,U −V⟩

= (u1 −v1)2 + (u2 −v2 )

2 + (u3 −v3)2 + (u4 −v4 )

2

= (2 −0)2 + ((−1)−9)2 + (3−4)2 + (1−6)2

= 4 +100 +1+ 25 = 130.

12

Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Polynomials) in P2?


Let P2 be the set of all polynomials of degree less than or equal to two. For two polynomials

we define

p(x)=a0x0 + a1x

1 + a2x2 ,

q(x) =b0x0 +b1x

1 +b2x2

rp,

rq ∈P2 ,

⟨rp,

rq⟩= a0b0 + a1b1 + a2b2 ,

rp = ⟨

rp,

rp⟩

12 = a2

0 + a21 + a2

2 , and

d(rp,

rq) =

rp −

rq = ⟨

rp −

rq,

rp −

rq⟩

12

= (a0 − b0 )2 + (a1 − b1)12 + (a3 − b3)

2 .


13

Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Polynomials) in P2? (Cont.)


Example 3: Find the inner product, norms, and distance for p and q in P2, where

⟨rp,

rq⟩= a0b0 + a1b1 + a2b2 = (8)(1) + (−3)(−2) + (5)(7) = 8 + 6 + 35 = 49,

rp = ⟨

rp,

rp⟩

12 = a0

2 + a21 + a2

2 = 82 + (−3)2 + 52 = 98 = 7 2,

rq = ⟨

rq,

rq⟩

12 = b0

2 + b12 + b2

2 = 12 + (−2)2 + 72 = 54 = 3 6, and

d(rp,

rq) =

rp −

rq = ⟨

rp −

rq,

rp −

rq⟩

12 = (a0 − b0 )2 + (a1 − b1)1

2 + (a3 − b3)2

= (8 −1)2 + ((−3) − (−2))2 + (5 − 7)2 = 49 +1+ 4 = 54 = 3 6.

p(x)=8−3x+ 5x2 and q(x) =1−2x+ 7x2 .

14

Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Functions) in C[a,b]?


⟨rf ,

rg⟩= f (x)g(x)dx

a

b

∫ ,

rf = ⟨

rf ,

rf ⟩

12 = [ f (x)]2 dx

a

b

∫⎛

⎝⎜⎞

⎠⎟

12

= f 2 (x)dxa

b

∫ ,

and d(rf ,

rg) =

rf −

rg = ⟨

rf −

rg,

rf −

rg⟩

12 = [ f (x) − g(x)]2 dx

a

b

∫⎛

⎝⎜⎞

⎠⎟

12

.

Let C[a,b] be the set of all continuous functions on [a,b].

For all f and g in C[a,b], we define


15

Rev.F09

How to Find an Inner Product, Norm, and Distance for Vectors (Functions) in C[a,b]? (Cont.)


Example 4: Let f = f(x) = 1 and g = g(x) = -x. Find the inner product, norms, and distance on C[2,4].

⟨rf ,

rg⟩= f (x)g(x)dx

2

4

∫ = (1)(−x)dx2

4

∫ = −x2

22

4

= (−8) − (−2) = −6,

rf = ⟨

rf ,

rf ⟩

12 = [ f (x)]2 dx

2

4

∫⎛

⎝⎜⎞

⎠⎟

12

= f 2 (x)dx2

4

∫ = dx2

4

∫ == x2

4= 2,

rg = ⟨

rg,

rg⟩

12 = [g(x)]2 dx

2

4

∫⎛

⎝⎜⎞

⎠⎟

12

= g2 (x)dx2

4

∫ = (−x)2 dx2

4

∫ ==x3

32

4

= 214

3,

and d(rf ,

rg) =

rf −

rg = ⟨

rf −

rg,

rf −

rg⟩

12 = [ f (x) − g(x)]2 dx

2

4

∫⎛

⎝⎜⎞

⎠⎟

12

= (1+ x)2 dx2

4

∫ = (1+ 2x + x2 )dx2

4

∫ = (x + x2 +x3

3)

2

4

= 72

3.

16

Rev.F09

The Cauchy-Schwarz Inequality


Cauchy-Schwarz Inequality:

for all f, g in an inner product space V, with the norm induced by

the inner product. We evaluate the Cauchy-Schwarz inequality

for the previous examples.

⟨rf ,

rg⟩ ≤

rf

rg

1. ⟨ru,

rv⟩ =1≤ 6 5 =

ru

rv from Example 1.

2 ⟨U,V⟩ =9 ≤ 15 133 = U V from Example 2.

3. ⟨rp,

rq⟩ =49 ≤(7 2)(3 6) =

rp

rq from Example 3.

4. ⟨rf ,

rg⟩ =6 ≤( 2)(2

143) =

rf

rg from Example 4.

17

Rev.F09

Orthogonality for Vectors (Functions) in C[a,b]


Example 5: Let f = f(x) = 1 and g = g(x) = ex. Find the inner

product on C[0,2].

Since the inner product is not zero, f and g are not orthogonal

functions in C[0,2].

Example 6: Let f = f(x) = 5 and g = g(x) = cos(x). Find the

inner product on C[0,π].

Since the inner product is zero, f and g are orthogonal functions

in C[0,π].

⟨rf ,

rg⟩= f (x)g(x)dx

0

2

∫ = (1)(ex )dx0

2

∫ = e2 − e0 = e2 −1 ≠ 0

⟨rf ,

rg⟩= f (x)g(x)dx

0

π

∫ = 5cos(x)dx0

π

∫ == 5 cos(x)dx0

π

∫= 5[sin(π ) − sin(0)] = 5[0 − 0] = 0

18

Rev.F09

Orthogonality for Vectors (Functions) in C[a,b] (Cont.)


Example 7: Let f = f(x) = 4x and g = g(x) = x2. Find the inner product on C[-1,1].

Since the inner product is zero, f and g are orthogonal functions in C[-1,1].

⟨rf ,

rg⟩= f (x)g(x)dx

−1

1

∫ = (4x)(x2 )dx−1

1

∫ == 4 x3 dx−1

1

∫ = x4( )−1

1= 1−1 = 0

19

Rev.F09

Inner Product and Orthogonality for Vectors (Functions) in C[a,b] (Cont.)


Example 8: Let f = f(x) = 1 and g = g(x) = sin(2x). Find the inner product on C[-π,π].

Since the inner product is zero, f and g are orthogonal functions in C[-π,π].

⟨rf ,

rg⟩= f (x)g(x)dx

−π

π

∫ = (1)(sin(2x))dx−π

π

∫ = sin(2x)dx−π

π

∫ = −1

2cos(2x)

−π

π

= −1

2cos(2π )⎡

⎣⎢⎤⎦⎥

− −1

2cos(−2π )⎡

⎣⎢⎤⎦⎥

= −1

2cos(2π ) +

1

2cos(2π ) = 0,

cos(2x) is even.


20

Rev.F09

The Generalized Theorem of Pythogoras


Generalized Theorem of Pythogoras: If f and g are orthogonal vectors in an inner product space V, then

Proof: Let f and g be orthogonal in V, then

Thus,rf +

rg

2=⟨

rf +

rg,

rf +

rg⟩ =⟨

rf ,

rf +

rg⟩ + ⟨

rg,

rf +

rg⟩

=⟨rf ,

rf ⟩ + ⟨

rf ,

rg⟩ + ⟨

rg,

rf ⟩ + ⟨

rg,

rg⟩

=rf

2+ 2⟨

rf ,

rg⟩ +

rg 2

=rf

2+

rg 2 .

⟨rf ,

rg⟩= 0.

21

Rev.F09

The Generalized Theorem of Pythogoras (Cont.)


The Generalized Theorem of Pythogoras

We evaluate the theorem for the previous examples. From example 6, we have

rf +

rg

2= [5 + cos(x)]2 dx

0

π

∫ = (25 + (2)(5)cos(x) + cos2 (x))dx0

π

∫

= 25dx0

π

∫ + 2 (5)cos(x)dx0

π

∫ + cos2 (x)dx0

π

∫

=rf

2+ 2⟨5,cos(x)⟩ +

rg 2 =

rf

2+ 2(0) +

rg 2 =

rf

2+

rg 2 .

22

Rev.F09



From Example 7, we have

Notice that we have used orthogonality to obtain our results without directly computing the integrals.

Next, we will compute the integrals directly to obtain the result.

rf +

rg

2= [4x+ x2 ]2 dx

−1

1

∫ = (16x2 + (2)(4x)x2 + x4 )dx−1

1

∫

= 16x2 dx−1

1

∫ + 2 4x3 dx−1

1

∫ + x4 dx−1

1

∫

=rf

2+ 2⟨4x,x2 ⟩ +

rg 2 =

rf

2+ 2(0) +

rg 2 =

rf

2+

rg 2 .

23

Rev.F09



Thus,

rf

2=⟨

rf ,

rf ⟩ = f 2 (x)dx

−1

1

∫ = [ f (x)]2 dx−1

1

∫ = (4x)(4x)dx−1

1

∫ ==16 x2 dx−1

1

∫ =(16)x3

3−1

1

=323

rg 2 =⟨

rg,

rg⟩ = g2 (x)dx

−1

1

∫ = [g(x)]2 dx−1

1

∫ = (x2 )(x2 )dx−1

1

∫ == x4 dx−1

1

∫ =x5

5−1

1

=25

rf +

rg

2=⟨

rf +

rg,

rf +

rg⟩ = [ f (x) + g(x)]2 dx

−1

1

∫ = (4x+ x2 )(4x+ x2 )dx−1

1

∫

= (16x2 + 8x3 + x4 )dx−1

1

∫ =16x3

3+8x4

4+

x5

5⎛

⎝⎜⎞

⎠⎟−1

1

=16615

=323+25=

rf

2+

rg 2 .

24

Rev.F09

How to Find the Cosine of the Angle Between Two Vectors in an Inner Product Space?


Example 9: Let ⁵ have the Euclidean inner product. ℜ

Find the cosine of the angle between

u = (2,1,0,-1,5) and v = (1,-1,-2,3,1).

cos(θ) =⟨ru,

rv⟩

ru

rv=

u1v1 +u2v2 +u3v3 + .u4v4 + +u5v5u12 +u2

2 +u32 +u4

2 +u52 v1

2 + v22 + v3

2 + v42 + v5

2

=(2)(1) + (1)(−1) + (0)(−2) + (−1)(3) + (5)(1)

22 +12 + 0 + (−1)2 + 52 12 + (−1)2 + (−2)2 + (3)2 +12

=3

4 31.


25

Rev.F09

How to Find the Cosine of the Angle Between Two Vectors in an Inner Product Space? (Cont.)


Example 10: Let P2 have the previous inner product.

Find the cosine of the angle between p and q.

cos(θ) =⟨rp,

rq⟩

rp

rq=

a0b0 + a1b1 + a2b2a02 + a1

2 + a22 b0

2 +b12 +b2

2

=(2)(−2) + (6)(−1) + (3)(4)

22 + 62 + 32 (−2)2 + (−1)2 + (4)2

=2

7 21.

rp =p(x) =a0x

0 + a1x1 + a2x

2 =2 + 6x+ 3x2 ,rq=q(x) =b0x

0 +b1x1 +b2x

2 =−2−x+ 4x2


26

Rev.F09

Quick Review on a Basis for a Vector Space and the Dimension of a Vector Space


Let be nonzero vectors in V. Then for , we have that

W = span(S) = {all linear combinations of } is a subspace of V.

If S contains some linearly dependent vectors, then the dim(W) < n.

If S is a linearly independent set of vectors, then S is a basis for the vector space W. W is a proper subspace of V, if dim(W) < dim(V). If dim(W) = dim(V) = n, then S is a basis for V.

A non-trivial vector space V always have two subspaces:

the trivial vector space, {0} and V.

rv1,

rv2 ,...,

rvn

S ={rv1,

rv2 ,...,

rvn}

rv1,

rv2 ,...,

rvn

27

Rev.F09

Quick Review on a Basis for a Vector Space and the Dimension of a Vector Space


So, we need two conditions for the set S to be a basis of a vector space, it must be a linearly independent set and it must span the vector space. The number of basis vectors in S is the dimension of the vector space.

For example: dim( ²ℜ )=2, dim( ³ℜ )=3, dim( ⁵ℜ )=5,

dim (P2)=3, dim(M22)=4, and dim(C[a,b]) is infinite.

28

Rev.F09

Rank, Row Rank, and Column Rank of a Matrix


The dimensions of the row space of A, column space of A, and nullspace of A are also called the row rank of A, column rank of A, and nullity of A, respectively.

The rank of A = the row rank of A = the column rank of A, and is the number of linearly independent columns in the matrix A.

The nullity of A = dim(null(A)) and is the number of free variables in the solution space.

29

Rev.F09

The Orthogonal Complement of a Subspace W


The orthogonal complement of W, is the set of all vectors in a finite dimensional inner product space V that are orthogonal to every vector in W.

Both W and are subspaces of V where

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

The only vector common to W and is 0, and W ⊥

W ⊥

W ⊥W ⊥

W =(W⊥ )⊥ .

30

Rev.F09

The Four Fundamental Matrix Spaces


Let A be an m x n matrix, then:

1.The null(A) and the row(A) are orthogonal complements in ⁿ with respect to the Euclidean inner product. ℜ

2.The null(AT) and the col(A) are orthogonal complements in ℜᵐ with respect to the Euclidean inner product.

Note that row(A) = col(AT) and col(A) = row(AT).

The four fundamental matrix spaces are:

• row(A) = col(AT),

• col(A) = row(AT),

• null(A), and

• null(AT).

31

Rev.F09

How to Find a Basis for the Orthogonal Complement of the Subspace of ⁿℜ Spanned by the Vectors?


Example 11: Find a basis for the orthogonal complement of the subspace of ⁿℜ spanned by the vectors v1, v2, and v3.

Let V = span({v1, v2, v3}). Then, V is a subspace of of ⁿℜ if v1, v2, v3 ∈ ⁿℜ . Let

From the last module, we know that the null(A) is the orthogonal complement to the row(A) = col(AT), and that the null(A) is the solution space of the homogeneous system, Ax = 0. Let v1, v2, v3 be the row vectors of the matrix A. Then, row(A) is the orthogonal complement to null(A) which is the solution space to Ax = 0. We shall use Gauss Elimination to reduce A to a row echelon form to find the basis for null(A).

rv1 =(1,−1,3),

rv2 =(5,−4,−4),

rv3 =(7,−6,2).


32

Rev.F09

How to Find a Basis for the Orthogonal Complement of the Subspace of ⁿℜ Spanned by the Vectors? (Cont.)


r1r2r3

157

−1−4−6

3−42

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=A

r1−5r1+ r2−7r1+ r3

100

−111

3−19−19

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

r1r2

−r2 + r3

100

−110

3−190

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Since the null(A) is the solution space of the homogeneous system, Ax = 0, the general solution of the system is:

x3 =s,x2 −19x3 =0,x2 −19s=0,x2 =19s.x1 −x2 + 3x3 =0,x1 −19s+ 3s=0,x1 =16s.

Note that column 1 and column 2, with the red leading 1’s, are linearly independent. The corresponding columns of A form a basis for the col(A). Then, dim(col(A)) = 2.


33

Rev.F09



The nonzero rows with the leading 1’s are linearly independent. These rows form a basis for row(A). Then, dim(row(A)) = 2.

Note that the row space and column space are both two dimensional, so rank(A) = 2. The rank(A) is the common dimension of the row space of A and the column space of A.

So, rank(A) = rowrank(A) = colrank(A) = dim(col(A)) =dim(row(A)) is always true for any m x n matrix A.

34

Rev.F09



Thus, the solution vector is:

and the vector

{w1} forms a basis for the null(A). In other words, {w1} = {(16,19,1)} forms a basis for the orthogonal complement of row(A).

x1

x2

x3

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

16s19ss

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=s16191

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

rw1 =

16191

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

35

Rev.F09



Since B = {w1} is a basis for null(A), span(B) = null(A), and

dim(null(A)) = 1 is the nullity of A. In this example, rank(A) is 2 and the nullity of A is 1, and the number of columns of A is 3.

For any m x n matrix A, rank(A) + nullity(A) is n, which is the number of columns of A.

36

Rev.F09

Let’s Review Some Equivalent Statements

If A is an n x n matrix, and if TA: ⁿ→ ⁿ is multiplication by A, ℜ ℜthan the following are equivalent. (See Theorem 6.2.7)

• A is invertible.• Ax = 0 has only the trivial solution • The reduced row-echelon form of A is In.• A is expressible as a product of elementary matrices. • Ax = b is consistent for every n x 1 matrix b. • Ax = b has exactly one solution for every n x 1 matrix b. • det(A) ≠ 0.• The range of TA is ⁿℜ .• TA is one-to-one.



37

Rev.F09

Let’s Review Some Equivalent Statements (Cont.)

• The column vectors of A are linearly independent.• The row vectors of A are linearly independent.• The column vectors of A span ⁿℜ . • The row vectors of A span ⁿℜ . • The column vectors of A form a basis for ⁿℜ . • The row vectors of A form a basis for ⁿℜ .• A has rank n = row rank n = column rank n = dim(col(A)).• A has nullity 0 = dim(null(A)).• The orthogonal complement of the nullspace of A,

null(A), is ⁿℜ .• The orthogonal complement of the row space of A,

row(A), is {0}.


38

Rev.F09

What have we learned?We have learned to:

1. Define and find the inner product, norm and distance in a given inner product space.

2. Find the cosine of the angle between two vectors and determine if the vectors are orthogonal in an inner product space.

3. Find the orthogonal complement of a subspace of an inner product space.

4. Find a basis for the orthogonal complement of a subspace of ⁿ ℜspanned by a set of row vectors.

5. Identify the four fundamental matrix spaces of an m x n matrix A, and know the column rank of A, the row rank of A, and the rank of A.

6. Know the equivalent statements of an n x n matrix A.


39

Rev.F09

Credit

Some of these slides have been adapted/modified in part/whole from the following textbook:

• Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition



1 mac 2103 module 10 lnner product spaces i. 2 rev.f09 learning objectives upon completing this...

Documents