Engineering Mathematics with Examples and Applications

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Engineering Mathematics-I
Eigenvalues and Eigenvectors
Prepared By- Prof. Mandar Vijay Datar
I2IT, Hinjawadi, Pune – 411057
http://www.isquareit.edu.in/

Definition
Definition 1: A nonzero vector x is an eigenvector (or characteristic
vector) of a square matrix A if there exists a scalar λ such that Ax = λx.
Then λ is an eigenvalue (or characteristic value) of A.
Note: The zero vector can not be an eigenvector but zero can be an
eigenvalue.
Example: Claim that is an eigen-vector for matrix
Solution:- Observe that For λ= 3
λx = Thus, λ = 3 is an eigenvalue of A and x is the
corresponding eigen vector.







1
1
x 







41
21
A




















3
3
1
1
41
21
Ax












3
3
1
1
3

Geometric interpretation
Let A be an n×n matrix, if X is an n×1 vector then Y=AX is
another n×1 vector. Thus A can be considered as a
transformation matrix that transforms vector X to vector Y
In general, a matrix multiplied to a vector changes both its
magnitude and direction. However, a matrix may operate on
certain vectors by changing only their magnitude, and leaving
their direction unchanged. Such vectors are the Eigenvectors
of the matrix.
If a matrix multiplied to an eigenvector changes its magnitude
by a factor, which is positive if its direction is unchanged and
negative if its direction is reversed. This factor is nothing but the
eigenvalue associated with that eigenvector.

Eigenvalues
Let x be an eigenvector of the matrix A. Then there must exist an
eigenvalue λ such that Ax = λx
or, equivalently, Ax - λx = 0 or (A – λI)x = 0
If we define a new matrix B = A – λI, then Bx = 0
If B has an inverse then x = B-10 = 0. But an eigenvector cannot
be zero. Thus, it follows that x will be an eigenvector of A if and
only if B does not have an inverse, or equivalently det(B)=0, or
det(A – λI) = 0
This is called the characteristic equation of A. Its roots are the
eigenvalues of A.

Eigenvalues: examples
Example 1: Find the eigenvalues of
Thus, two eigenvalues: 5,  1
Note: The roots of the characteristic equation can be repeated. That is, λ1 = λ2
=…= λk. If that happens, the eigenvalue is said to be of multiplicity k.
Example 2: Find the eigen values of







32
41
A
)1)(5(54
8)3)(1(
32
41
IA
2


















131
111
322
A

0)3)(1)(2(
0652IA
131
111
322
IA
23






Example continued…..
Therefore, λ = -2, 1, 3 are eigenvalues of A.

Eigenvectors
To each distinct eigenvalue of a matrix A there will be at least one
corresponding eigenvector which can be obtained by solving the appropriate
set of homogenous equations.
If λi is an eigenvalue then the corresponding eigenvector xi is the solution of
system (A – λiI)xi = 0
Example 1 (cont.): Let, λ =5. Consider the following
system
X
00
11
X
22
44
X)I5A( 














Solving the above system, we get eigenvector corresponding to eigen value λ

0t,
1
1
t
x
x
tx,tx0xx
2
1
2121














x













00
21
42
42
I)1(A:1
0s,
1
2
s
x
x
2
1
2 











x

Example 2 (cont.): Find the eigenvectors of
Recall that λ = -2 is an eigenvalue of A
Solve the homogeneous linear system represented by
Let
The eigenvectors of  = -2 are of the form






























 

0
0
0
x
x
x
131
131
324
)I)2(A(
3
2
1
x tx2 



































14
1
11
t
t14
t
t11
x
x
x
3
2
1
x













131
111
322
A

Properties of Eigenvalues and Eigenvectors
Definition: The trace of a matrix A, denoted by tr(A), is the sum of
the elements present on the main diagonal of matrix A.
Property 1: The sum of the eigenvalues of a matrix equals the
trace of the matrix.
Property 2: A matrix is singular if and only if it has a zero
eigenvalue.
Property 3: The eigenvalues of an upper (or lower) triangular
matrix are the elements on the main diagonal.
Property 4: If λ is an eigenvalue of A and A is invertible, then 1/λ
is an eigenvalue of matrix A-1.

Property 5: If λ is an eigenvalue of A then kλ is an eigenvalue of
kA where k is any arbitrary scalar.
Property 6: If λ is an eigenvalue of A then λk is an eigenvalue of
Ak for any positive integer k.
Property 8: If λ is an eigenvalue of A then λ is an eigenvalue of AT.
Property 9: The product of the eigenvalues (counting multiplicity)
of a matrix equals the determinant of the matrix.

Important Results-
Theorem: Eigenvectors corresponding to distinct eigenvalues are
linearly independent.
Theorem: If λ is an eigenvalue of multiplicity k of an n  n matrix
A then the number of linearly independent eigenvectors of A
associated with λ is given by m = n - rank(A- λI).
Furthermore, 1 ≤ m ≤ k.

Engineering Mathematics with Examples and Applications

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