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Transcript of Maths>>Eigenvalues and eigenvectors
Eigenvalues and
Eigenvectors
Eigenvalues and Eigenvectors
• If A is an n x n matrix and λ is a scalar for which Ax = λx has a nontrivial solution x ∈ ⁿ, then ℜ λ is an eigenvalue of A and x is a corresponding eigenvector of A.
Ax=λx=λIx (AλI)x=0
• The matrix (AλI ) is called the characteristic matrix of a where I is the Unit matrix.
• The equation det (AλI )= 0 is called characteristic equationof A and the roots of this equation are called the eigenvalues of the matrix A. The set of all eigenvectors is called the eigenspace of A corresponding to λ. The set of all eigenvalues of a is called spectrum of A.
Characteristic Equation• If A is any square matrix of order n, we can form
the matrix , where is the nth order unit matrix.
• The determinant of this matrix equated to zero,
• i.e.,
is called the characteristic equation of A.
IλA
0
λa...aa............a...λaaa...aλa
λA
nnn2n1
2n2221
1n1211
I
• On expanding the determinant, we get
• where k’s are expressible in terms of the elements a
• The roots of this equation are called Characteristic roots
or latent roots or eigen values of the matrix A.
•X = is called an eigen vector or latent vector
0k...λkλkλ1)( n2n
21n
1nn
ij
4
2
1
x...xx
5
Properties of Eigen Values:1. The sum of the eigen values of a matrix is the
sum of the elements of the principal diagonal.
2. The product of the eigen values of a matrix A is equal to its determinant.
3. If is an eigen value of a matrix A, then 1/ is the eigen value of A1 .
4. If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value.
6
PROPERTY 1: If λ1, λ2,…, λn are the eigen values of A, then
i. k λ1, k λ2,…,k λn are the eigen values of the matrix kA, where k is a non – zero scalar.
ii. are the eigen values of the inverse matrix A1.
iii. are the eigen values of Ap, where p is any positive integer.
n21 λ1,...,
λ1,
λ1
pn
p2
p1 λ...,,λ,λ
Algebraic & Geometric Multiplicity
• If the eigenvalue λ of the equation det(AλI)=0 is repeated n times then n is called the algebraic multiplicity of λ. The number of linearly independent eigenvectors is the difference between the number of unknowns and the rank of the corresponding matrix A λI and is known as geometric multiplicity of eigenvalue λ.
CayleyHamilton Theorem
• Every square matrix satisfies its own characteristic
equation.
• Let A = [aij]n×n be a square matrix then,
nnnn2n1n
n22221
n11211
a...aa................a...aaa...aa
A
Let the characteristic polynomial of A be (λ)Then,
The characteristic equation is
11 12 1n
21 22 2n
n1 n2 nn
φ(λ) = A  λIa  λ a ... a
a a  λ ... a=
... ... ... ...a a ... a  λ
 A  λI = 0
Note 1: Premultiplying equation (1) by A1 , we have
n n1 n20 1 2 n
n n1 n20 1 2 n
We are to prove that
p λ +p λ +p λ +...+p = 0
p A +p A +p A +...+p I= 0 ...(1)
I
n1 n2 n3 10 1 2 n1 n
1 n1 n2 n30 1 2 n1
n
0 = p A +p A +p A +...+p +p A1A = [p A +p A +p A +...+p I]p
This result gives the inverse of A in terms of (n1) powers of A and is considered as a practical method for the computation of the inverse of the large matrices.
Note 2: If m is a positive integer such that m > n then any positive integral power Am of A is linearly expressible in terms of those of lower degree.
Verify Cayley – Hamilton theorem for the matrix
A = . Hence compute A1 .
Solution: The characteristic equation of A is
211121
112
tion)simplifica(on049λ6λλor
0λ211
1λ2111λ2
i.e.,0λIA
23
Example 1:
To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1)
Now,
222121212221212222
211121
112
655565
556
655565
556
211121
112
211121
112
23
2
AAA
A
A3 6A2 +9A – 4I = 0 =  6 + 9
4
=
This verifies Cayley – Hamilton theorem.
222121212221212222
655565
556
211121
112
100010001
0000000000
15
Now, pre – multiplying both sides of (1) by A1 , we have
A2 – 6A +9I – 4 A1 = 0 => 4 A1 = A2 – 6 A +9I
311131113
41
311131113
100010001
9211121
1126
655565
5564
1
1
A
A
16
Given find Adj A by using Cayley –
Hamilton theorem.
Solution: The characteristic equation of the given matrix A is
113110121
A
tion)simplifica(on035λ3λλor
0λ1131λ1012λ1
i.e.,0λIA
23
Example 2:
17
By Cayley – Hamilton theorem, A should satisfyA3 – 3A2 + 5A + 3I = 0
Pre – multiplying by A1 , we get A2 – 3A +5I +3A1 = 0
339330363
3A
146223452
113110121
113110121
A.A ANow,
(1)...5I)3A(A31A
2
21
18
AAAAdj.
AAAdj.Athat,knowWe
173143110
31
500050005
339330363
146223452
31AFrom(1),
1
1
1
19
173143110
AAdj.
173143110
313)(AAdj.
3113110121
ANow,
Similarity of Matrix
• If A & B are two square matrices of order n then B is said to be similar to A, if there exists a nonsingular matrix P such that,B= P1AP
1. Similarity matrices is an equivalence relation.2. Similarity matrices have the same determinant.3. Similar matrices have the same characteristic polynomial
and hence the same eigenvalues. If x is an eigenvector corresponding to the eigenvalue λ, then P1x is an eigenvector of B corresponding to the eigenvalue λ where B= P1AP.
Diagonalization
• A matrix A is said to be diagonalizable if it is similar to diagonal matrix.
• A matrix A is diagonalizable if there exists an invertible matrix P such that P1AP=D where D is a diagonal matrix, also known as spectral matrix. The matrix P is then said to diagonalize A of transform A to diagonal form and is known as modal matrix.
22
Reduction of a matrix to Diagonal Form
• If a square matrix A of order n has n linearly independent eigen vectors then a matrix B can be found such that B1AB 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.
23
Reduce the matrix A = to diagonal form by
similarity transformation. Hence find A3.
Solution: Characteristic equation is
=> λ = 1, 2, 3 Hence eigenvalues of A are 1, 2, 3.
300120
211
0
λ3001λ20
21λ1
Example:
24
Corresponding to λ = 1, let X1 = be the eigen
vector then
3
2
1
xxx
001
kX
x0x,kx02x0xx02xx
000
xxx
200110
2100X)I(A
11
3211
3
32
32
3
2
1
1
25
Corresponding to λ = 2, let X2 = be the eigen
vector then,
3
2
1
xxx
011
kX
xkx,kx0x
0x02xxx
000
xxx
100100
211
0X)(A
22
32221
3
3
321
3
2
1
2
0,
I2
26
Corresponding to λ = 3, let X3 = be the eigen
vector then,
3
2
1
xxx
223
kX
xkx,kx
0x02xxx
000
xxx
000110
212
0X)(A
33
13332
3
321
3
2
1
3
3
2
23,
2
I3
k
x
27
Hence modal matrix is
2100110
2111
MMAdj.M
100220122
MAdj.
2M
200210
311M
1
28
Now, since D = M1AM
=> A = MDM1
A2 = (MDM1) (MDM1)
= MD2M1 [since M1M = I]
300020001
200210
311
300120
211
2100110
2111
AMM 1
29
Similarly, A3 = MD3M1
=
A3 =
270019803271
2100110
2111
2700080001
200210
311
Orthogonally Similar Matrices
• If A & B are two square matrices of order n then B is said to be orthogonally similar to A, if there exists orthogonal matrix P such that B= P1AP
Since P is orthogonal,P1=PT
B= P1AP=PTAP1. A real symmetric of order n has n mutually orthogonal real
eigenvectors.2. Any two eigenvectors corresponding to two distinct
eigenvalues of a real symmetric matrix are orthogonal.
31
Diagonalise the matrix A = by means of an
orthogonal transformation.Solution: Characteristic equation of A is
204060402
66,2,λ0λ)16(6λ)λ)(2λ)(6(2
0λ204
0λ6040λ2
Example :
32
I
1
1 2
3
1
1
2
3
1 3
2
1 3
1 1 2 3 1
1 1
xwhenλ = 2,let X = x be theeigenvector
x
then (A +2 )X = 0
4 0 4 x 00 8 0 x = 04 0 4 x 0
4x + 4x = 0 ...(1)8x = 0 ...(2)
4x + 4x = 0 ...(3)x = k ,x = 0,x = k
1X = k 0
1
33
2
2I
0
1
2
3
1
2
3
1 3
1 3
1 3 2
2 2 3
xwhenλ = 6,let X = x be theeigenvector
x
then (A  6 )X = 0
4 0 4 x 00 0 x = 04 0 4 x 0
4x + 4x = 04x  4x = 0
x = x and x isarbitraryx must be so chosen that X and X are orthogonal among th
.1
emselvesand also each is orthogonal with X
34
2 3
3 1
3 2
3
1 αLet X = 0 and let X = β
1 γ
Since X is orthogonal to Xα  γ = 0 ...(4)
X is orthogonal to Xα+ γ = 0 ...(5)
Solving (4)and(5), we get α = γ = 0 and β is arbitrary.0
Taking β =1, X = 10
1 1 0Modal matrix is M = 0 0 1
1 1
0
35
The normalised modal matrix is1 1 02 2
N = 0 0 11 1 02 21 10  1 1 02 2 2 0 4 2 21 1D =N'AN = 0 0 6 0 0 0 12 2
4 0 2 1 1 00 1 02 2
2 0 0D = 0 6 0 which is the required diagonal matrix
0 0 6.
36
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.
Quadratic Forms
37
In n – variables x1,x2,…,xn, the general quadratic form is
In the expansion, the coefficient of xixj = (bij + bji).
n
1j
n
1ijiijjiij bbwhere,xxb
).b(b21awherexxaxxb
baandaawherebb2aSuppose
jiijijji
n
1j
n
1iijji
n
1j
n
1iij
iiiijiijijijij
38
Hence every quadratic form can be written as
getweform,matrixinformsquadraticofexamplessaidabovethewritingNow
.x,...,x,xXandaAwheresymmetric,alwaysisAmatrixthethatso
AX,X'xxa
n21ij
ji
n
1j
n
1iij
)(by2hxy ax (i) 22
ii
yx
bhha
y][xby2hxy ax (i) 22
39
wzyx
dnmlncgfmgbhlfha
wzyx
2nzw 2myw 2lxw zx 2f 2gyz2hxy dw2 cz by ax (iii)
zyx
cgfgbhfha
zyx2fzx 2gyz 2hxy cz by ax (ii)
222
222
40
Two Theorems On Quadratic Form
Theorem(1): A quadratic form can always be expressed with respect to a given coordinate system as
where A is a unique symmetric matrix.Theorem2: Two symmetric matrices A and B represent
the same quadratic form if and only if B=PTAP where P is a nonsingular matrix.
AxxY T
Nature of Quadratic FormA 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.
42
Find the nature of the following quadratic formsi. x2 + 5y2 + z2 + 2xy + 2yz + 6zxii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2xySolution: iii. The matrix of the quadratic form is
113151311
A
Example :
43
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.
311151
113A
Linear Transformation of a Quadratic Form
44
• Let X’AX be a quadratic form in n variables and let X = PY ….. (1) where P is a non – singular matrix, be the non – singular transformation.
• From (1), X’ = (PY)’ = Y’P’ and hence
X’AX = Y’P’APY = Y’(P’AP)Y
= Y’BY …. (2)
where B = P’AP.
45
Therefore, Y’BY is also a quadratic form in n variables. Hence it is a linear transformation of the quadratic form X’AX under the linear transformation X = PY and B = P’AP.
Note. (i) Here B = (P’AP)’ = P’AP = B
(ii) ρ(B) = ρ(A)
Therefore, A and B are congruent matrices.
46
Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form.
OrDiagonalise the quadratic form 3x2 + 3z2 + 4xy +
8xz + 8yz by linear transformations and write the linear transformation.
OrReduce the quadratic form 3x2 + 3z2 + 4xy + 8xz +
8yz into the sum of squares.
Example:
47
Solution: The given quadratic form can be written as X’AX where X = [x, y, z]’ and the symmetric matrix
A =
Let us reduce A into diagonal matrix. We know tat A = I3AI3.
344402423
100010001
344402423
100010001
344402423
48
21 31OperatingR ( 2 / 3),R ( 4 / 3)(for A onL.H.S.andpre factor on R.H.S.),weget
3 2 4 1 0 0 1 0 04 4 20 1 0 A 0 1 03 3 3
0 0 14 7 40 0 13 3 3
10001034
321
A
1034
0132
001
37
340
34
340
423
getweR.H.S),onfactorpostandL.H.S.onA(for4/3)(C2/3),(C Operating 3121
49
10001034
321
A
112
0132
001
10034
340
003
getwe(1),R Operating 32
APP'1,343,Diagor
100110
2321
A
112
0132
001
100
0340
003
getwe(1),C Operating 32
50
The canonical form of the given quadratic form is
Here ρ(A) = 3, index = 1, signature = 1 – (2) = 1.Note: In this problem the nonsingular
transformation which reduces the given quadratic form into the canonical form is X = PY.
i.e.,
3
2
1
112
0132
001
yyy
zyx
23
22
21
3
2
1
321
yy343y
yyy
100
0340
003yyyAP)Y(P'Y'
THANK YOU