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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 ∈ ℜ...
Characteristic Equation
• If A is any square matrix of order n, we can form the
matrix , where is the nth order unit matri...
• On expanding the determinant, we get
• where k’s are expressible in terms of the elements a
• The roots of this equation...
5
Properties of Eigen Values:-
1. The sum of the eigen values of a matrix is the
sum of the elements of the principal diag...
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,...
Algebraic & Geometric Multiplicity
• If the eigenvalue λ of the equation det(A-λI)=0
is repeated n times then n is called ...
Cayley-Hamilton Theorem
• Every square matrix satisfies its own characteristic
equation.
• Let A = [aij]n n be a square ma...
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 - λ...
Note 1:- Premultiplying equation (1) by A-1 , we
have
n n-1 n-2
0 1 2 n
n n-1 n-2
0 1 2 n
We are to prove that
p λ +p λ +p...
This result gives the inverse of A in terms of
(n-1) powers of A and is considered as a practical
method for the computati...
Verify Cayley – Hamilton theorem for the matrix
A = . Hence compute A-1 .
Solution:- The characteristic equation of A is
2...
To verify Cayley – Hamilton theorem, we have to
show that A3 – 6A2 +9A – 4I = 0 … (1)
Now,
222121
212221
212222
211
121
11...
A3 -6A2 +9A – 4I = 0
= - 6 + 9
-4
=
This verifies Cayley – Hamilton theorem.
222121
212221
212222
655
565
556
211
121
112
...
15
Now, pre – multiplying both sides of (1) by A-1 , we
have
A2 – 6A +9I – 4 A-1 = 0
=> 4 A-1 = A2 – 6 A +9I
311
131
113
4...
16
Given find Adj A by using Cayley –
Hamilton theorem.
Solution:- The characteristic equation of the given
matrix A is
11...
17
By Cayley – Hamilton theorem, A should satisfy
A3 – 3A2 + 5A + 3I = 0
Pre – multiplying by A-1 , we get
A2 – 3A +5I +3A...
18
AAAAdj.
A
AAdj.
Athat,knowWe
173
143
110
3
1
500
050
005
339
330
363
146
223
452
3
1
AFrom(1),
1
1
1
19
173
143
110
AAdj.
173
143
110
3
1
3)(AAdj.
3
113
110
121
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 no...
Diagonalization
• A matrix A is said to be diagonalizable if it is
similar to diagonal matrix.
• A matrix A is diagonaliza...
22
Reduction of a matrix to Diagonal Form
• If a square matrix A of order n has n linearly
independent eigen vectors then ...
23
Reduce the matrix A = to diagonal form by
similarity transformation. Hence find A3.
Solution:- Characteristic equation ...
24
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
...
25
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
10...
26
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
...
27
Hence modal matrix is
2
1
00
11-0
2
1-
11
M
MAdj.
M
1-00
220
122-
MAdj.
2M
200
21-0
311
M
1
28
Now, since D = M-1AM
=> A = MDM-1
A2 = (MDM-1) (MDM-1)
= MD2M-1 [since M-1M = I]
300
020
001
200
21-0
311
300
120
211
2...
29
Similarly, A3 = MD3M-1
=
A3 =
2700
19-80
327-1
2
1
00
11-0
2
1
11
2700
080
001
200
21-0
311
Orthogonally Similar Matrices
• If A & B are two square matrices of order n then B is said to
be orthogonally similar to A...
31
Diagonalise the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
204
060...
32
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 betheeigenvector
x
then (A + 2 )X = 0
4 0 4 x 0
0 8 0 ...
33
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 betheeigenvector
x
then (A -6 )X = 0
-4 0 4 x 0
0 0 x = 0
...
34
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 = ...
35
The normalised modal matrix is
1 1
0
2 2
N = 0 0 1
1 1
- 0
2 2
1 1
0 - 1 1
02 2 2 0 4 2 2
1 1
D =N'AN = 0 0 6 0 0 0 1
2...
36
DEFINITION:-
A homogeneous polynomial of second degree
in any number of variables is called a quadratic
form.
For examp...
37
In n – variables x1,x2,…,xn, the general quadratic form
is
In the expansion, the co-efficient of xixj = (bij + bji).
n
...
38
Hence every quadratic form can be written as
getweform,matrixin
formsquadraticofexamplessaidabovethewritingNow
.x,...,x...
39
w
z
y
x
dnml
ncgf
mgbh
lfha
wzyx
2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii)
z
y
x
cgf
gbh
fha
zyx2fzx2gyz2hxyczbyax(ii)
222...
40
Two Theorems On Quadratic Form
Theorem(1): A quadratic form can always be expressed
with respect to a given coordinate ...
Nature of Quadratic Form
A real quadratic form X’AX in n variables is said to
be
i. Positive definite if all the eigen val...
42
Find the nature of the following quadratic forms
i. x2 + 5y2 + z2 + 2xy + 2yz + 6zx
ii. 3x2 + 5y2 + 3z2 – 2yz + 2zx – 2...
43
The eigen values of A are -2, 3, 6.
Two of these eigen values being positive and
one being negative, the given quadratr...
Linear Transformation of a
Quadratic Form
44
• Let X’AX be a quadratic form in n- variables and let
X = PY ….. (1) where P...
45
Therefore, Y’BY is also a quadratic form in n-
variables. Hence it is a linear transformation of
the quadratic form X’A...
46
Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical
form.
Or
Diagonalise the quadratic form 3x2 + 3z2 + 4xy +
8xz + 8yz b...
47
Solution:- The given quadratic form can be
written as X’AX where X = [x, y, z]’ and the
symmetric matrix
A =
Let us red...
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 0
4 4 2
0 1 0 A 0 1 0...
49
100
010
3
4
3
2
1
A
112
01
3
2
001
100
3
4
3
4
0
003
getwe(1),ROperating 32
APP'1,
3
4
3,Diagor
100
110
2
3
2
1
A
112
0...
50
The canonical form of the given quadratic form is
Here ρ(A) = 3, index = 1, signature = 1 – (2) = -1.
Note:- In this pr...
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Maths-->>Eigenvalues and eigenvectors

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Maths-->>Eigenvalues and eigenvectors

  1. 1. Eigenvalues and Eigenvectors
  2. 2. 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 ∈ ℜⁿ, then λ is an eigenvalue of A and x is a corresponding eigenvector of A. Ax=λx=λIx (A-λI)x=0 • The matrix (A-λI ) is called the characteristic matrix of a where I is the Unit matrix. • The equation det (A-λI )= 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.
  3. 3. 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...λaa a...aλa λA nnn2n1 2n2221 1n1211 I
  4. 4. • 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)( n 2n 2 1n 1 nn ij 4 2 1 x ... x x
  5. 5. 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 A-1 . 4. If is an eigen value of an orthogonal matrix, then 1/ is also its eigen value.
  6. 6. 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 A-1. iii. are the eigen values of Ap, where p is any positive integer. n21 λ 1 ,..., λ 1 , λ 1 p n p 2 p 1 λ...,,λ,λ
  7. 7. Algebraic & Geometric Multiplicity • If the eigenvalue λ of the equation det(A-λI)=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.
  8. 8. Cayley-Hamilton 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...aa a...aa A
  9. 9. 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 - λI a - λ a ... a a a - λ ... a = ... ... ... ... a a ... a - λ | A - λI|=0
  10. 10. Note 1:- Premultiplying equation (1) by A-1 , we have n n-1 n-2 0 1 2 n n n-1 n-2 0 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) In-1 n-2 n-3 -1 0 1 2 n-1 n -1 n-1 n-2 n-3 0 1 2 n-1 n 0 =p A +p A +p A +...+p +p A 1 A =- [p A +p A +p A +...+p I] p
  11. 11. This result gives the inverse of A in terms of (n-1) 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.
  12. 12. Verify Cayley – Hamilton theorem for the matrix A = . Hence compute A-1 . Solution:- The characteristic equation of A is 211 121 112 tion)simplifica(on049λ6λλor 0 λ211 1λ21 11λ2 i.e.,0λIA 23 Example 1:-
  13. 13. To verify Cayley – Hamilton theorem, we have to show that A3 – 6A2 +9A – 4I = 0 … (1) Now, 222121 212221 212222 211 121 112 655 565 556 655 565 556 211 121 112 211 121 112 23 2 AAA A
  14. 14. A3 -6A2 +9A – 4I = 0 = - 6 + 9 -4 = This verifies Cayley – Hamilton theorem. 222121 212221 212222 655 565 556 211 121 112 100 010 001 0 000 000 000
  15. 15. 15 Now, pre – multiplying both sides of (1) by A-1 , we have A2 – 6A +9I – 4 A-1 = 0 => 4 A-1 = A2 – 6 A +9I 311 131 113 4 1 311 131 113 100 010 001 9 211 121 112 6 655 565 556 4 1 1 A A
  16. 16. 16 Given find Adj A by using Cayley – Hamilton theorem. Solution:- The characteristic equation of the given matrix A is 113 110 121 A tion)simplifica(on035λ3λλor 0 λ113 1λ10 1-2λ1 i.e.,0λIA 23 Example 2:-
  17. 17. 17 By Cayley – Hamilton theorem, A should satisfy A3 – 3A2 + 5A + 3I = 0 Pre – multiplying by A-1 , we get A2 – 3A +5I +3A-1 = 0 339 330 363 3A 146 223 452 113 110 121 113 110 121 A.AANow, (1)...5I)3A(A 3 1 A 2 21-
  18. 18. 18 AAAAdj. A AAdj. Athat,knowWe 173 143 110 3 1 500 050 005 339 330 363 146 223 452 3 1 AFrom(1), 1 1 1
  19. 19. 19 173 143 110 AAdj. 173 143 110 3 1 3)(AAdj. 3 113 110 121 ANow,
  20. 20. 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 non-singular matrix P such that, B= P-1AP 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 P-1x is an eigenvector of B corresponding to the eigenvalue λ where B= P-1AP.
  21. 21. 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 P-1AP=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. 22. 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 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.
  23. 23. 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. 300 120 211 0 λ-300 1λ-20 21λ1- Example:-
  24. 24. 24 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
  25. 25. 25 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
  26. 26. 26 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
  27. 27. 27 Hence modal matrix is 2 1 00 11-0 2 1- 11 M MAdj. M 1-00 220 122- MAdj. 2M 200 21-0 311 M 1
  28. 28. 28 Now, since D = M-1AM => A = MDM-1 A2 = (MDM-1) (MDM-1) = MD2M-1 [since M-1M = I] 300 020 001 200 21-0 311 300 120 211 2 1 00 11-0 2 1 11 AMM 1
  29. 29. 29 Similarly, A3 = MD3M-1 = A3 = 2700 19-80 327-1 2 1 00 11-0 2 1 11 2700 080 001 200 21-0 311
  30. 30. 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= P-1AP Since P is orthogonal, P-1=PT B= P-1AP=PTAP 1. 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. 31. 31 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 Example :-
  32. 32. 32 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 betheeigenvector 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
  33. 33. 33 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 betheeigenvector 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
  34. 34. 34 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
  35. 35. 35 The normalised modal matrix is 1 1 0 2 2 N = 0 0 1 1 1 - 0 2 2 1 1 0 - 1 1 02 2 2 0 4 2 2 1 1 D =N'AN = 0 0 6 0 0 0 1 2 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 .
  36. 36. 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. 37. 37 In n – variables x1,x2,…,xn, the general quadratic form is In the expansion, the co-efficient of xixj = (bij + bji). n 1j n 1i jiijjiij bbwhere,xxb ).b(b 2 1 awherexxaxxb baandaawherebb2aSuppose jiijijji n 1j n 1i ijji n 1j n 1i ij iiiijiijijijij
  38. 38. 38 Hence every quadratic form can be written as getweform,matrixin formsquadraticofexamplessaidabovethewritingNow .x,...,x,xXandaAwhere symmetric,alwaysisAmatrixthethatso AX,X'xxa n21ij ji n 1j n 1i ij y x bh ha y][xby2hxyax(i) 22
  39. 39. 39 w z y x dnml ncgf mgbh lfha wzyx 2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii) z y x cgf gbh fha zyx2fzx2gyz2hxyczbyax(ii) 222 222
  40. 40. 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 non-singular matrix. AxxY T
  41. 41. Nature of 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.
  42. 42. 42 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 Example :-
  43. 43. 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. 311 151 113 A
  44. 44. 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. 45. 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. 46. 46 Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form. Or Diagonalise the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz by linear transformations and write the linear transformation. Or Reduce the quadratic form 3x2 + 3z2 + 4xy + 8xz + 8yz into the sum of squares. Example:-
  47. 47. 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. 344 402 423 100 010 001 344 402 423 100 010 001 344 402 423
  48. 48. 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 0 4 4 2 0 1 0 A 0 1 0 3 3 3 0 0 1 4 7 4 0 0 1 3 3 3 100 010 3 4 3 2 1 A 10 3 4 01 3 2 001 3 7 3 4 0 3 4 3 4 0 423 getweR.H.S),onfactorpostandL.H.S.onA(for 4/3)(C2/3),(COperating 3121
  49. 49. 49 100 010 3 4 3 2 1 A 112 01 3 2 001 100 3 4 3 4 0 003 getwe(1),ROperating 32 APP'1, 3 4 3,Diagor 100 110 2 3 2 1 A 112 01 3 2 001 100 0 3 4 0 003 getwe(1),COperating 32
  50. 50. 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 non-singular transformation which reduces the given quadratic form into the canonical form is X = PY. i.e., 3 2 1 112 01 3 2 001 y y y z y x 2 3 2 2 2 1 3 2 1 321 yy 3 4 3y y y y 100 0 3 4 0 003 yyyAP)Y(P'Y'
  51. 51. THANK YOU

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