linear algebra
DESCRIPTION
maths 2TRANSCRIPT
Prepared by:-
1)Hardik Sonaiya (140110707014)
2)Ketan Gohil (140110707015)
Guidance by:-
Prof. Mukesh Joshi
Diagonalisation of a matrix A is the
process of reduction A to a diagonal form.
If A is related to D by a similarity transformation,
such that D = M-1AM then A is reduced to the
diagonal matrix D through modal matrix M. D is
also called spectral matrix of A.
2
If a square matrix A of order n has n linearly
independent eigen vectors then a matrix B can
be found such that B-1AB is a diagonal matrix.
Note:- The matrix B which diagonalises A is called
the modal matrix of A and is obtained by
grouping the eigen vectors of A into a square
matrix.
3
Similarity of matrices:-
A square matrix B of order n is said to be a similar to a square matrix A of order n if
B = M-1AM for some non singular matrix M.
This transformation of a matrix A by a non – singular matrix M to B is called a similarity transformation.
Note:- If the matrix B is similar to matrix A, then B
has the same eigen values as A.
4
Reduce the matrix A = to diagonal form by
similarity transformation. Hence find A3.
Solution:- Characteristic equation is
=> λ = 1, 2, 3 Hence eigen values of A are 1, 2, 3.
300
120
211
5
0
λ-300
1λ-20
21λ1-
Example:-
Corresponding to λ = 1, let X1 = be the eigen
vector then
3
2
1
x
x
x
0
0
1
kX
x0x,kx
02x
0xx
02xx
0
0
0
x
x
x
200
110
210
0X)I(A
11
3211
3
32
32
3
2
1
1
6
Corresponding to λ = 2, let X2 = be the eigen
vector then,
3
2
1
x
x
x
0
1-
1
kX
x-kx,kx
0x
0x
02xxx
0
0
0
x
x
x
100
100
211-
0X)(A
22
32221
3
3
321
3
2
1
2
0,
I2
7
Corresponding to λ = 3, let X3 = be the eigen
vector then,
3
2
1
x
x
x
2
2-
3
kX
xk-x,kx
0x
02xxx
0
0
0
x
x
x
000
11-0
212-
0X)(A
33
13332
3
321
3
2
1
3
3
2
2
3,
2
I3
k
x
8
Hence modal matrix is
2
100
11-02
1-11
M
MAdj.M
1-00
220
122-
MAdj.
2M
200
21-0
311
M
1
9
300
020
001
200
21-0
311
300
120
211
2
100
11-02
111
AMM 1
10
Now, since D = M-1AM
=> A = MDM-1
A2 = (MDM-1) (MDM-1)
= MD2M-1 [since M-1M =
I]
Similarly, A3 = MD3M-1
=
A3 =
2700
19-80
327-1
2
100
11-02
111
2700
080
001
200
21-0
311
11
A square matrix A with real elements is said
to be orthogonal if AA’ = I = A’A.
But AA-1 = I = A-1A, it follows that A is orthogonal if
A’ = A-1.
Diagonalisation by orthogonal transformation
is possible only for a real symmetric matrix.
12
If A is a real symmetric matrix then eigen
vectors of A will be not only linearly independent but
also pairwise orthogonal.
If we normalise each eigen vector and use
them to form the normalised modal matrix N then it
can be proved that N is an orthogonal matrix.
13
The similarity transformation M-1AM = D
takes the form N’AN = D since N-1 = N’ by a
property of orthogonal matrix.
Transforming A into D by means of the
transformation N’AN = D is called as orthogonal
reduction or othogonal transformation.
Note:- To normalise eigen vector Xr, divide each
element of Xr, by the square root of the sum of the
squares of all the elements of Xr.
14
Diagonalise the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
204
060
402
66,2,λ
0λ)16(6λ)λ)(2λ)(6(2
0
λ204
0λ60
40λ2
15
Example :-
I
1
1 2
3
1
1
2
3
1 3
2
1 3
1 1 2 3 1
1 1
x
whenλ = -2,let X = x be theeigenvector
x
then (A +2 )X = 0
4 0 4 x 0
0 8 0 x = 0
4 0 4 x 0
4x + 4x = 0 ...(1)
8x = 0 ...(2)
4x + 4x = 0 ...(3)
x = k ,x = 0,x = -k
1
X = k 0
-1
16
2
2I
0
1
2
3
1
2
3
1 3
1 3
1 3 2
2 2 3
x
whenλ = 6,let X = x be theeigenvector
x
then (A - 6 )X = 0
-4 0 4 x 0
0 0 x = 0
4 0 -4 x 0
4x + 4x = 0
4x - 4x = 0
x = x and x isarbitrary
x must be so chosen that X and X are orthogonal among th
.1
emselves
and also each is orthogonal with X
17
2 3
3 1
3 2
3
1 α
Let X = 0 and let X = β
1 γ
Since X is orthogonal to X
α - γ = 0 ...(4)
X is orthogonal to X
α+ γ = 0 ...(5)
Solving (4)and(5), we get α = γ = 0 and β is arbitrary.
0
Taking β =1, X = 1
0
1 1 0
Modal matrix is M = 0 0 1
-1 1
0
18
The normalised modal matrix is
1 10
2 2N = 0 0 1
1 1- 0
2 2
1 10 - 1 1
02 2 2 0 4 2 21 1
D =N'AN = 0 0 6 0 0 0 12 2
4 0 2 1 1- 00 1 0
2 2
-2 0 0
D = 0 6 0 which is the required diagonal matrix
0 0 6
.
19
DEFINITION:- DEFINITION:-
A homogeneous polynomial of second degree in any number of variables is called a quadratic form.
For example,
ax2 + 2hxy +by2
ax2 + by2 + cz2 + 2hxy + 2gyz + 2fzx and
ax2 + by2 + cz2 + dw2 +2hxy +2gyz + 2fzx + 2lxw + 2myw + 2nzw
are quadratic forms in two, three and four variables.
20
In n – variables x1,x2,…,xn, the general quadratic form
is
In the expansion, the co-efficient of xixj = (bij + bji).
n
1j
n
1ijiijjiij bbwhere,xxb
21
).b(b2
1awherexxaxxb
baandaawherebb2aSuppose
jiijijji
n
1j
n
1iijji
n
1j
n
1iij
iiiijiijijijij
Hence every quadratic form can be written as
getweform,matrixin
formsquadraticofexamplessaidabovethewritingNow
.x,...,x,xXandaAwhere
symmetric,alwaysisAmatrixthethatso
AX,X'xxa
n21ij
ji
n
1j
n
1iij
22
y
x
bh
hay][xby2hxy ax (i) 22
w
z
y
x
dnml
ncgf
mgbh
lfha
wzyx
2nzw 2myw 2lxw zx 2f 2gyz2hxy dw2 cz by ax (iii)
z
y
x
cgf
gbh
fha
zyx2fzx 2gyz 2hxy cz by ax (ii)
222
222
23
1.11 NATURE OF A QUADRATIC FORM
A real quadratic form X’AX in n variables is said to be
i. Positive definite if all the eigen values of A > 0.ii. Negative definite if all the eigen values of A < 0.iii. Positive semidefinite if all the eigen values of A 0
and at least one eigen value = 0.iv. Negative semidefinite if all the eigen values of A 0 and at least one eigen value = 0.v. Indefinite if some of the eigen values of A are + ve
and others – ve.
24
Find the nature of the following quadratic forms
i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx
ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xy
Solution:-
i. The matrix of the quadratic form is
113
151
311
A
25
Example :-
The eigen values of A are -2, 3, 6.Two of these eigen values being positive and one being negative, the given quadratric form is indefinite.
ii. The matrix of the quadratic form is
The eigen values of A are 2, 3, 6. All these eigen values being positive, the given quadratic form is positive definite.
311
151
113
A
26
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