Power Method for Eigenvalues

J.P.
5/13/2022

Introduction - The Power Method

  • We use the power method to find dominant eigenvalues and their corresponding eigenvectors for square matrices.
  • The power method is an iterative process

Introduction: What are Eigenvectors and Eigenvalues?

  • An eigenvector is a nonzero vector that when we multiply the eigenvector to the original matrix, we get a scalar multiple of the vector.
  • Let \( A \) be our matrix, \( \mathbf{v} \) be the eigenvector, and \( \lambda \) be the scalar multiple, then: \[ A\mathbf{v} = \lambda \mathbf{v}. \]
  • The scalar multiple \( \lambda \) is the eigenvalue.
  • A matrix may have multiple eigenvalues. The largest of these is known as the dominant eigenvalue.
  • The dominant eigenvectors are the vectors that corresponds with the dominant eigenvalue.

Introduction - What are Eigenvectors and Eigenvalues?

  • In linear algebra, eigenvalues are typically found by solving the characteristic equation of an \( n \times n \) matrix \( A \). \[ det(\lambda I - A) = \lambda ^n + c_{n-1}\lambda ^{n-1}+ ...+ c_0 = 0 \]

Description of Method

  • We use an iterative process to approximate the dominant eigenvalue and eigenvector of a square matrix.
  • We begin with an initial guess, \( \mathbf{v}_0 \), for the eigenvector.
  • To get the next vector in the iteration, we use: \[ \mathbf{v}_{n+1} = A\mathbf{v}_{n}. \]
  • The estimated eigenvector is the vector \( \mathbf{v} \) with a scalar multiple divided out.
  • As we iterate, we should be getting closer and closer to the true dominant eigenvector.
  • If the matrix does not have a dominant eigenvector, the Power Method will not converge.
  • To get the corresponding eigenvalue \( \lambda \), we use the Rayleigh quotient: \[ \lambda = \frac{A \mathbf{v}^T*\mathbf{v}}{\mathbf{v}^T\mathbf{v}} \]

Example

We will use the Power Method to approximate the dominant eigenvalue and eigenvector for the matrix: \[ A = \begin{bmatrix} 2& -12 \\ 1& -5 \\ \end{bmatrix} \] We start with an initial guess for the dominant eigenvector: \[ \mathbf{v}_0= \begin{bmatrix} 1\\ 1\\ \end{bmatrix} \] Using the method described, we obtain the following: \[ \mathbf{v}_1 = A\mathbf{v}_0 = \begin{bmatrix} 2& -12\\ 1& -5\\ \end{bmatrix} \begin{bmatrix} 1\\ 1\\ \end{bmatrix} = \begin{bmatrix} -10\\ -4\\ \end{bmatrix} \Rightarrow -4 \begin{bmatrix} 2.50 \\ 1.00\\ \end{bmatrix} \] And after repeating 6 times we get: \[ \mathbf{v}_6 = A\mathbf{v}_5 = \begin{bmatrix} 2& -12\\ 1& -5\\ \end{bmatrix} \begin{bmatrix} -280\\ -94\\ \end{bmatrix} = \begin{bmatrix} 598\\ 190\\ \end{bmatrix} \Rightarrow 190 \begin{bmatrix} 2.99 \\ 1.00\\ \end{bmatrix} \]

Octave Code


A = [0.49, 0.02, 0.22; 0.02, 0.28, 0.20; 0.22, 0.20,  0.40]  
# Defining matrix A
v = [1;1;1]         #initial guess
v = A*v;
[M, I] = max(v);  # Finding location of 
v_scaled = v/v(I) # largest value in first iteration

itr = 1;

while itr < 11     # loop starts
  v = A*v;   # calculating new vector
  v_scaled = v/v(I) #scaling vector
  itr = itr+1;
end

Octave Code


#Rayleigh quotient
s1 = A*v_scaled; 
s2 = s1*v_scaled';
s3 = v_scaled'*v_scaled;
eig = s2/s3;

fprintf('The dominant eigenvalue is %d and the dominant eigenvector is \n', eig)
v_scaled

Octave Output

A =

   0.490000   0.020000   0.220000
   0.020000   0.280000   0.200000
   0.220000   0.200000   0.400000

v =

   1
   1
   1

The dominant eigenvalue is 0.72 and the dominant eigenvector is 
v_scaled =

   0.9999
   0.5001
   1.0000

Iteration Output

plot of chunk unnamed-chunk-5

Discussion of Results

After 10 iterations the Power Method gave an estimated dominant eigenvector of \[ \begin{bmatrix} .9999 \\ 0.5001\\1.0000\\ \end{bmatrix} \]

and an estimated dominant eigenvalue of \( 0.72 \) for the matrix

\[ A = \begin{bmatrix} 0.49 0.02 0.22\\ 0.02 0.28 0.20 \\ 0.22 0.20 0.40 \\ \end{bmatrix}. \]

Discussion of Results

Using analytical methods from linear algebra the dominant eigenvector is \[ \begin{bmatrix} 1.0 \\ 0.5\\1.0\\ \end{bmatrix} \]

with a corresponding eigenvalue of 0.72.

This results in a 0% error for the dominant eigenvalue and a 0.0022% error for the dominant eigenvector.

Conclusion

The Power Method is a quick and efficient numerical method of finding dominant eigenvectors and eigenvalues for a square matrix. With 10 iterations we had 0.0022% error for a \( 3 \times 3 \) matrix, and with more iterations the percent error will decrease. The downfall is that if you can only find a dominant eigenvector and eigenvalue.

References