For GTU Students. Subject: VCLA (2110015)

1. Eigen Value And Eigen
Vector
GUIDED BY: MANSI K. DESAI

2. Group Members:-
SR NO. NAME ENROLLMENT NUMBER
1. RANA PAYAL MAHESHBHAI 151100106072
2. PATEL RUTVIJ GANESHBHAI 151100106073
3. SAVALIYAAKSHAY JAYANTILAL 151100106074
4. TANDEL PAYALBENVITHTHALBHAI 151100106075
5. TANDEL SNEHAL MAHENDRABHAI 151100106076
6. VAGHELA SAHIL PREMJIBHAI 151100106077
7. EHSANULLAHAYDIN 151100106078
8. SHUAIB KOHEE 151100106079
9. WAHEDULLAH EHSAS 151100106080

3. 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 equation of 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.

4. 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.
0
λa...aa
............
a...λaa
a...aλa
λA
nnn2n1
2n2221
1n1211




 I

5. • 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
 












4
2
1
...
x
x
x

6. 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.


7. 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.

8. 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.

9. 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















10. 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

11.  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)
Note 1:- Premultiplying equation (1) by A-1 , we
have
I

n-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

12. 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.

13. Example 1:-
Verify Cayley – Hamilton theorem for the matrix
A = . Hence compute A-1 .
Solution:-The characteristic equation of A is





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






211
121
112
tion)simplifica(on049λ6λλor
0
λ211
1λ21
11λ2
i.e.,0λIA
23







14. 
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
222121
212221
212222
211
121
112
655
565
556
655
565
556
211
121
112
211
121
112
23
2
AAA
A
To verify Cayley – Hamilton theorem, we have to
show that A3 – 6A2 +9A – 4I = 0 … (1)
Now,

15. A3 -6A2 +9A – 4I = 0
= -6 + 9
-4
=
This verifies Cayley – Hamilton theorem.









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


222121
212221
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655
565
556
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211
121
112

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100
010
001
0
000
000
000












16. 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



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
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311
131
113
4
1
311
131
113
100
010
001
9
211
121
112
6
655
565
556
4
1
1
A
A

17. Example 2:-
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







18. 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





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339
330
363
3A
146
223
452
113
110
121
113
110
121
A.AANow,
(1)...5I)3A(A
3
1
A
2
21-

19. 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
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20. 
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173
143
110
AAdj.
173
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110
3
1
3)(AAdj.
3
113
110
121
ANow,

21. Similarity of Matrix
▪ IfA & 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.

22. 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.

23. 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 ofA into a square
matrix.

24. Example:-
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-

25. Corresponding to λ = 1, let X1 = be the eigen
vector then 









3
2
1
x
x
x















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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

26. Corresponding to λ = 2, let X2 = be the eigen
vector then,










3
2
1
x
x
x















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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

27. Corresponding to λ = 3, let X3 = be the eigen
vector then, 

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






3
2
1
x
x
x
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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

28. Hence modal matrix is

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2
1
00
11-0
2
1-
11
M
MAdj.
M
1-00
220
122-
MAdj.
2M
200
21-0
311
M
1

29. 












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300
020
001
200
21-0
311
300
120
211
2
1
00
11-0
2
1
11
AMM 1
Now, since D = M-1AM
=> A = MDM-1
A2 = (MDM-1) (MDM-1)
= MD2M-1 [since M-1M = I]

30. Similarly, A3 = MD3M-1
=
A3 =

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
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
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

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
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






2700
19-80
327-1
2
1
00
11-0
2
1
11
2700
080
001
200
21-0
311

31. 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.

32. Diagonalises the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
32
Example :-
204
060
402
66,2,λ
0λ)16(6λ)λ)(2λ)(6(2
0
λ204
0λ60
40λ2







33. I
 
 
 
  
     
     
     
          


 
   
  
1
1 2
3
1
1
2
3
1 3
2
1 3
1 1 2 3 1
1 1
x
when λ = -2,let X = x be the eigen vector
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

34. 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

35.    
   
   
      


 
 
 
  

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

36.  
 
 
 
 
 
  
 
  
    
    
    
        
    
 
 
 
  
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
.

37. Quadratic Forms
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

38. In n – variables x1,x2,…,xn, the general quadratic form
is
In the expansion, the co-efficient of xixj = (bij + bji).
38
 

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


   

39. 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

40.  
 















































w
z
y
x
dnml
ncgf
mgbh
lfha
wzyx
2nzw2myw2lxwzx2f2gyz2hxydw2czbyax(iii)
z
y
x
cgf
gbh
fha
zyx2fzx2gyz2hxyczbyax(ii)
222
222

41. 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


42. 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 ofA > 0.
ii. Negative definite if all the eigen values of A < 0.
iii. Positive semi definite if all the eigen values ofA 0 and at least one
eigen value = 0.
iv. Negative semi definite if all the eigen values of
A 0 and at least one eigen value = 0.
v. Indefinite if some of the eigen values ofA are + ve and others – ve.

43. 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
43
Example :-











113
151
311
A

44. 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

45. Linear Transformation of a Quadratic
Form
▪ 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.

46. 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.

47. Reduce 3x2 + 3z2 + 4xy + 8xz + 8yz into canonical form.
Or
Diagonalises 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.
47
Example:-

48. 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

49.  

   
   
    
           
      
    
   
21 31OperatingR ( 2 / 3),R ( 4 / 3)
(for A onL.H.S.andpre factor on R.H.S.), we get
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

50. 






































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















































51. 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.,
 
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'

























































3
2
1
112
01
3
2
001
y
y
y
z
y
x

52. Thank you 