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1. EIGEN VALUES AND
EIGEN VECTORS
Continuous Assessment -1
Presented By: SatyabrataMondal
Department: CSE
UniversityRoll: 12000123082
Subject: MathematicsIA
Subject code: BS M101
Semester: 1st
College:Dr.B.C. Roy EngineeringCollege

2. INTRODUCTION
• 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. Determining Eigenvalues
● Vector equation
◦ Ax = x (A-)x = 0
◦ A- is called the characteristic matrix
● Non-trivial solutions exist if and only if:
◦ This is called the characteristic equation
● Characteristic polynomial
◦ nth-order polynomial in 
◦ Roots are the eigenvalues {1, 2, …, n}
= 0
−
an1 an2 ann
a12
a1n
a2n
⁝
a11 −
a21 a22
⁝
−
⁝ ⋱
det(A − I) =

4. Eigenvalue Example
● Characteristic matrix


     3 − 4 −
0 1 
0
=
1− 2
− 4
3
2 
− 
1
A− I =
1
● Characteristic equation
A −I = (1−)(−4 − ) − (2)(3) = 2
+3 −10 = 0
● Eigenvalues: 1 = -5, 2 = 2

5. Eigenvalue Properties
● Eigenvalues of A and AT are equal
● Singular matrix has at least one zero eigenvalue
● Eigenvalues of A-1: 1/1, 1/2, …, 1/n
● Eigenvalues of diagonal and triangular matrices
are equal to the diagonal elements
● Trace n
j =1
● Determinant
Tr ( A) =   j
n
A =  j
j=1

6. Determining Eigenvectors
● Eigenvectors determined up to scalar
multiple
● Distinct eigenvalues
◦ Produce linearly independent eigenvectors
● Repeated eigenvalues
◦ Produce linearly dependent eigenvectors
◦ Procedure to determine eigenvectors more complex
(see text)
◦ Will demonstrate in Matlab
● First determine eigenvalues: {1, 2, …, n}
● Then determine eigenvector corresponding
to each eigenvalue:
(A − I)x = 0  (A − k I)xk = 0

7. Eigenvector Example
● Eigenvalues
● Determine eigenvectors: Ax = x
● Eigenvector for 1 = 2
 


 −3
1 1
3x1 + x2 = 0  0.9487 
● Eigenvector for 1 = -5
6x1 + 2x2 = 0
x =
−0.3162
or x =
 1 
 


 1
0.4472
2 2
x =
0.8944
or x =
2
3x1 −6x2 =0
−x1 +2x2 =0
2
3
 
− 4  = 2
2  1 = −5
A =
1

(1− )x1 + 2x2 = 0
3x1 −(4 + )x2 = 0
x1 + 2x2 = x1
3x1 − 4x2 = x2

8. Cayley-Hamilton Theorem



 
 
2 1 2 2 2 n
n 1 n 2 n n
● Let the characteristic polynomial of A be
 (λ)then,
φ ( λ ) = A - λ I
a 1 2 a 1 n
 a 1 1 - λ
 a a
= 
. . . . . . . . .
 a a
. . .
a - λ . . .
. . .
. . . a - λ

The characteristic equation
| A -λI|= 0

9. ● Verify Cayley – Hamilton theorem for
the matrix A= . Hence
compute A-1 .
●Solution:- The characteristic equation
of A is



2 
−1
1
−1
−1 2
−1 1 
2
(on simplification)
or λ3
−6λ2
+9λ−4 = 0
2−λ −1 1
A−λI = 0 i.e., −1 2−λ −1 = 0
1 −1 2−λ

10. 
22

−2
1

 
2
 
21
−1
=−21

6

1


5
6 −5 52 −1 1 2.2 −22 −21

6 −5
−1 2 22
−5 −1 −21

6

 
2
 
5
−1
 =−5 −5


2

1


1
−1
−1
2 2 6
−1 −1 −5
2 −1 12 −1 1 6 −5 5
A3
=A2
A=−5
A2
=−1
To verify Cayley – Hamilton theorem, we have to
show that A3 – 6A2 +9A – 4I = 0 … (1)

11.  A3 -6A2 +9A – 4I = 0
− 21
 22 − 22 − 21

= − 21 22 
 21 − 21 22 

6 

 5
− 5
− 5
 6 − 5 5 
6 
− 5
- 6 + 9  
2 


1
−1
−1
 2 −1 1 
 2
−1
 
0 1
0

-4 0
1 0 0
1
0

0
= 0

0
0 0 0
0 0 = 0
0
This verifies Cayley – Hamilton theorem

12. References
• Engineering Mathematics
• https://prezi.com
• https://byjus.com/maths/eigen-values/