Numerical Methods - Power Method for Eigen values

Numerical Methods
Power Method for Eigen values
Dr. N. B. Vyas
Department of Mathematics,
Atmiya Institute of Technology & Science,
Rajkot (Gujarat) - INDIA
niravbvyas@gmail.com
Dr. N. B. Vyas Numerical Methods Power Method for Eigen values

Eigen values and Eigen vectors by iteration
Power Method
Power method is particularly useful for estimating numerically
largest or smallest eigenvalue and its corresponding eigenvector.

Power Method
The intermediate (remaining) eigenvalues can also be found.

Power Method
The power method, which is an iterative method, can be used
when

Power Method
when
(i) The matrix A of order n has n linearly independent eigenvectors.

Power Method
when
(ii) The eigenvalues can be ordered in magnitude as
|λ1| > |λ2| ≥ |λ3| ≥ . . . ≥ |λn|

Power Method
when
|λ1| > |λ2| ≥ |λ3| ≥ . . . ≥ |λn|
When this ordering is adopted, the eigenvalue λ1 with the
greatest magnitude is called the dominant eigenvalue of the
matrix A

Power Method
when
|λ1| > |λ2| ≥ |λ3| ≥ . . . ≥ |λn|
When this ordering is adopted, the eigenvalue λ1 with the
greatest magnitude is called the dominant eigenvalue of the
matrix A
And the remaining eigenvalues λ2, λ3, . . . , λn are called the
subdominant eigenvalues of A.

Power Method: Working rules for determining largest eigenvalue.

Let A = [aij] be a matrix of order n × n.

We start from any vector x0(= 0) with n components such that
Ax0 = x

Ax0 = x
In order to get a convergent sequence of eigenvectors
simultaneously scaling method is adopted.

Ax0 = x
In which at each stage each components of the resultant
approximate vector is to be divided by its absolutely largest
component.

Ax0 = x
component.
Then use the scaled vector in the next step.

Ax0 = x
component.
Then use the scaled vector in the next step.
This absolutely largest component is known as numerically
largest eigenvalue.

Accordingly x in eq -(1) can be scaled by dividing each of its
components by absolutely largest component of it. Thus
Ax0 = x = λ1x1; x1 is the scaled vector of x

Now scaled vector x1 is to be used in the next iteration to obtain
Ax1 = x = λ2x2

Ax1 = x = λ2x2
Proceeding in this way, ﬁnally we get Axn = λn+1xn+1; where
n = 0, 1, 2, 3, ... Where λn+1 is the numerically largest eigenvalue
upto desired accuracy and xn+1 is the corresponding eigenvector.

Ax1 = x = λ2x2
NOTE : The initial vector x0 is usually taken as a vector with
all components equal to 1.

Ax1 = x = λ2x2
NOTE : The initial vector x0 is usually taken as a vector with
all components equal to 1.
Characteristic: The main advantage of this method is its
simplicity. And it can handle sparse matrices too large to store
as a full square array. Its disadvantage is its possibly slow
convergence.

Power Method: Determining smallest eigenvalue.
If λ is the eigenvalue of A, then the reciprocal
1
λ
is the eigenvalue
of A−1.
The reciprocal of the largest eigenvalue of A−1 will be the
smallest eigenvalue of A.

Example
Ex: Use power method to estimate the largest eigen value and the
corresponding eigen vector of A =
3 −5
−2 4

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1
=
8
−6

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1
=
8
−6
= 8
1
−0.75
= 8x2

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1
=
8
−6
= 8
1
−0.75
= 8x2
Ax2 =
3 −5
−2 4
1
−0.75

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1
=
8
−6
= 8
1
−0.75
= 8x2
Ax2 =
3 −5
−2 4
1
−0.75
=
6.75
−5

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1
=
8
−6
= 8
1
−0.75
= 8x2
Ax2 =
3 −5
−2 4
1
−0.75
=
6.75
−5
= 6.75
1
−0.7407
= 6.75x3

Example
Sol.: Let A =
3 −5
−2 4
and x0 =
1
1
Ax0 =
3 −5
−2 4
1
1
=
−2
2
= −2
1
−1
= −2x1
Ax1 =
3 −5
−2 4
1
−1
=
8
−6
= 8
1
−0.75
= 8x2
Ax2 =
3 −5
−2 4
1
−0.75
=
6.75
−5
= 6.75
1
−0.7407
= 6.75x3
∴ largest eigen value is 6.7015 and the corresponding eigen vector is
1
−0.7403

Example
1 2
3 4

Example
2 3
5 4

Example
4 2
1 3

Example


2 −1 0
−1 2 −1
0 −1 2



Example


3 −1 0
−1 2 −1
0 −1 3



Numerical Methods - Power Method for Eigen values

More Related Content

What's hot (20)

Similar to Numerical Methods - Power Method for Eigen values (20)

More from Dr. Nirav Vyas (20)

Recently uploaded (20)

Numerical Methods - Power Method for Eigen values