It is the topic of maths-2 in the second semester of engineering !! I hope it is useful and satisfactory !!

1. Eigenvalues
and
Eigenvectors

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. 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. • 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
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
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. 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. 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. 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. 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. 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. 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. 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. 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
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
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
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
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
173
143
110
AAdj.
173
143
110
3
1
3)(AAdj.
3
113
110
121
ANow,

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. 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
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
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
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
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
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
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
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
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. 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
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
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
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
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
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
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
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
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
w
z
y
x
dnml
ncgf
mgbh
lfha
wzyx
2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii)
z
y
x
cgf
gbh
fha
zyx2fzx2gyz2hxyczbyax(ii)
222
222

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. 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
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
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. 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
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
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
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
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
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
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. THANK YOU