EigenValue and EigenVector

Let’s assume a square matrix A

The characteristic equation,

| A – λ*I | = 0

(where I is an identity matrix)

After calculating the values of λs we attempt to find eigenvectors for corresponding eigenvalues like this

For eigenvalue, λ = λ₁

A*x = λ₁*I*x (where, x is an unknown vector)

Or, (A - λ₁*I)*x = 0

The value of x is the corresponding eigenvector of λ₁

Power Method for Dominant Eigenvalue

Let λ₁, λ₂, λ₃, and λ_n be the eigenvalues of an n X n matrix A. λ₁ is called the dominant eigenvalue of A if

| λ₁| > | λ_i |, i = 2, 3, ... , n

The eigenvectors corresponding to λ₁ are called dominant eigenvectors of A.

Procedure

1. Choose an n X n matrix

The number of rows and columns should be the same (or matrix dimension mismatched)

2. Like the Jacobi and Gauss-Seidel methods, the power method for approximating eigenvalues is iterative. First, we assume that matrix A has a dominant eigenvalue with corresponding dominant eigenvectors. Then we choose an initial approximation x₀ of one of the

dominant eigenvectors of A. This initial approximation must be a nonzero vector in Rⁿ

Finally, we form the sequence given by

x₁ = Ax₀

x₂ = Ax₁ = A(Ax₀) = A²x₀

x₃ = Ax₂ = A(A²x₀) = A³x₀

_{. . .}

x_n = Ax_n-1 = A(A^n-1x₀) = Aⁿx₀

(In the above, x₁ denotes the value of vector x at the first iteration and so on)

Compare the updated value of x with its previous value (obtained from the previous iteration)

For large powers of k, and by properly scaling this sequence, we will see that we obtain a good approximation of the dominant eigenvector of A.

3. Repeat the iteration process until convergence

4. The formula for finding the corresponding eigenvalue from eigenvector x.

If x is an eigenvector of A, then its corresponding eigenvalue is given by

λ = (Ax.x / x.x)

5. If they do not converge even after many iterations (maybe after 1000 iterations), then

Entered matrix has no dominant eigenvalue

Example

A = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}$

We begin with an initial nonzero approximation of

x₀ = $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

We then obtain the following approximations

x₁ = Ax₀ = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ = $\begin{bmatrix} - 10 \\ - 4 \end{bmatrix}$ = -4$\begin{bmatrix} 2.50 \\ 1.00 \end{bmatrix}$

x₂ = Ax₁ = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} - 10 \\ - 4 \end{bmatrix}$ = $\begin{bmatrix} 28 \\ 10 \end{bmatrix}$ = 10$\begin{bmatrix} 2.80 \\ 1.00 \end{bmatrix}$

x₃ = Ax₂ = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} 28 \\ 10 \end{bmatrix}$ = $\begin{bmatrix} - 64 \\ - 22 \end{bmatrix}$ = -22$\begin{bmatrix} 2.91 \\ 1.00 \end{bmatrix}$

x₄ = Ax₃ = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} - 64 \\ - 22 \end{bmatrix}$ = $\begin{bmatrix} 136 \\ 46 \end{bmatrix}$ = 46$\begin{bmatrix} 2.96 \\ 1.00 \end{bmatrix}$

x₅ = Ax₄ = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} 136 \\ 46 \end{bmatrix}$ = $\begin{bmatrix} - 280 \\ - 94 \end{bmatrix}$ = -94$\begin{bmatrix} 2.98 \\ 1.00 \end{bmatrix}$

x₆ = Ax₅ = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} - 280 \\ - 94 \end{bmatrix}$ = $\begin{bmatrix} 568 \\ 190 \end{bmatrix}$ = 190$\begin{bmatrix} 2.99 \\ 1.00 \end{bmatrix}$

Note that the approximations in Example appear to be approaching scalar multiples of $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$

So, the obtained dominant eigenvector from the above iterations is

x = $\begin{bmatrix} 3 \\ 1 \end{bmatrix}$

Now, we’ll find the corresponding eigenvalue from the obtained eigenvector

Formula

If x is an eigenvector of A, then its corresponding eigenvalue is given by

λ = (Ax.x / x.x)

Ax = $\begin{bmatrix} 2 & - 12 \\ 1 & - 5 \end{bmatrix}\begin{bmatrix} 2.99 \\ 1.00 \end{bmatrix}$ = $\begin{bmatrix} - 6.02 \\ - 2.01 \end{bmatrix}$

Then, Ax.x = $\begin{bmatrix} - 6.02 \\ - 2.01 \end{bmatrix}\begin{bmatrix} 2.99 \\ 1.00 \end{bmatrix}$ = -20.0 (approx.)

And x.x = $\begin{bmatrix} 2.99 \\ 1.00 \end{bmatrix}\begin{bmatrix} 2.99 \\ 1.00 \end{bmatrix}$ = $\begin{bmatrix} 2.99 \\ 1.00 \end{bmatrix}$ = 9.94 (approx.)

So, the corresponding eigenvalue, λ = (-20.0 / 9.94) = -2 (approx.)