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I don't really get it, when the method converges..

The formula I have says that if the eigenvalue with the greatest absolute value has the algebraic multiplicity of 1, and is strictly greater than all other eigenvalues, which means if:

|lambda 1| > |lambda 2| >= |lambda 3| >= .... >= |lambda n|

then the method converges.

And if there are many eigenvalues with the same greatest absolute value, then xi can loop between 2 eigenvectors and not converge.

Now, I have this matrix:

matrix A

And I'm trying to iterate with the Rayleigh-Quotient-method, which is based on the power method, with starting value x_0 = (1, 0) and assumed eigenvalue near lambda_0 = 4.

I end up after 2nd iteration with lambda_2 = 5.0583 and x_2 = (0.4759, 0.8795)

And 3rd iteration with lambda_3 = 4.99999 and x_3 = (-0.4475, -15.495)

if I try the 4th iteration with these values, I get x_4 = (-0.4472, -0.8943) and lambda_4 = 4.999889.

What can I say about: 1) the convergence? 2) the absolute value greatest eigenvalue? 3) the eigenvector?

It seems that it's converging to Eigenvalue = 5, and eigenvector x = (-0.4472, -0.8943). Is it right or I'm wrong?

Thank you so much for any hints! I am really confused!

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  • $\begingroup$ You can verify it yourself. Multiply x with matrix. How much does it differ from 5x? $\endgroup$ Commented Jul 14, 2019 at 23:23

1 Answer 1

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What can I say about: 1) the convergence?

It converges when there is a dominant eigenvalue. It has trouble converging when they are very close together, which isn't the case here.

2) the absolute value greatest eigenvalue?

The matrix is small so you can check it by hand.

$$ A = \begin{pmatrix} 1 & 2 \\ 4 & 3 \end{pmatrix} $$

then

$$ \det(A- \lambda I) = \det(\begin{pmatrix} 1-\lambda & 2 \\ 4 & 3-\lambda \end{pmatrix}) = (1-\lambda)(3-\lambda) - 8 =\lambda^{2} - 4 \lambda - 5 $$

Then

$$ \lambda^{2} - 4\lambda -5 = (\lambda +1)(\lambda-5) \implies \lambda_{1} = 5 , \lambda_{2} = -1$$

3) the eigenvector?

To get the eigenvector we plug that in

$$ A-5 I = \begin{pmatrix} -4 & 2 \\ 4 & -2 \end{pmatrix}$$

Then we have that $v_{1} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ so

$$ q_{1} = \begin{pmatrix} \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}}\end{pmatrix}$$

I created a small program for this...

import  numpy as np
A = np.array([[1,2],[4,3]])
xk = np.array([[1],[0]])


def iterate(A,x):
    rho = np.dot(A,x)
    #lambda_i
    nk = np.max(np.absolute(rho))
    # v_i
    xk = 1/nk* rho

    return rho, nk, xk

n = 55

for i in range(n):
    if (i % 5 == 0):
        rho, nk, xk = iterate(A,xk)
        print("The {i}'th lambda iterate is:{lambda_i}".format(i=i+1,lambda_i =nk))
        print("The {i}'th eig_vec iterate is:{vec_i}".format(i=i+1,vec_i =xk))

If you use it, you should have

$$ x_{1} = \begin{pmatrix} \frac{1}{4} \\ 1 \end{pmatrix}$$

$$ x_{2} = \begin{pmatrix} .5625 \\ 1 \end{pmatrix}$$

$$ x_{3} = \begin{pmatrix} .4880 \\ 1 \end{pmatrix}$$

$$ x_{3} = \begin{pmatrix} .5024 \\ 1 \end{pmatrix}$$

$$ x_{4} = \begin{pmatrix} .49952 \\ 1 \end{pmatrix}$$

$$ x_{5} = \begin{pmatrix} .50009 \\ 1 \end{pmatrix}$$

after 50 iterations you'll have

$$ x_{50} =\begin{pmatrix} .5 \\ 1 \end{pmatrix} $$

or

$$ v_{1} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} $$

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  • $\begingroup$ Thank you for your help!! 1) what is v1 and what is q1? Is v1 the eigenvector to lambda_1? 2) But this is not using the power method. I have to use the power method to determine the absolute value greatest eigenvalue and the eigenvector. Clarifying my question 1: convergence: I mean is the power method converging or diverging? $\endgroup$
    – ZelelB
    Commented Jul 14, 2019 at 22:02
  • $\begingroup$ I'll change (add to) this one moment. $v_{1}$ is the solution you get but isn't orthonormal. $\endgroup$
    – user3417
    Commented Jul 14, 2019 at 22:06
  • $\begingroup$ Thanks so much!! and q1 is the eigenvector, right? Im writing my last exam in my degree, and this chapter is very important and confuses me a lot! Many thanks for your support here!! $\endgroup$
    – ZelelB
    Commented Jul 14, 2019 at 22:07
  • $\begingroup$ And I think you mean sqrt(5) not sqrt(3), right? $\endgroup$
    – ZelelB
    Commented Jul 14, 2019 at 22:10
  • $\begingroup$ yes, that is correct $\endgroup$
    – user3417
    Commented Jul 14, 2019 at 22:11

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