# Matrix Eigenvectors

• Apr 25th 2010, 08:04 PM
Laydieofsorrows
Matrix Eigenvectors
A is a 3 x 3 matrix with eigenvectors v1 = [1 0 0], v2 = [1 1 0], v3 = [1 1 1] (all vertical) corresponding to eigenvalues x1 = -1/3, x2 = 1/3, x3 = 1, respectively and x = [2 1 2].

Find (A^k)x. What happens as k approaches infinity?

I don't even know where to begin on this one.

Thanks for the help!
• Apr 25th 2010, 10:09 PM
eigenvex
First use the eigenvalue equation (Av = lambda*v) to backsolve for the values in A. You'll also have to use the property that the trace of a matrix (sum of the diagonal values) is also equal to the sum of that matrix's eigenvalues.

Once you have A, you'll see it is an upper-triangular matrix with values <= 1....and you can see what will happen if you raise this matrix to a high power. (if you can't see it, just type it into matlab and do A^5, A^6, ... until you see what is happening)
• Apr 26th 2010, 01:07 AM
Defunkt
Quote:

Originally Posted by Laydieofsorrows
A is a 3 x 3 matrix with eigenvectors v1 = [1 0 0], v2 = [1 1 0], v3 = [1 1 1] (all vertical) corresponding to eigenvalues x1 = -1/3, x2 = 1/3, x3 = 1, respectively and x = [2 1 2].

Find (A^k)x. What happens as k approaches infinity?

I don't even know where to begin on this one.

Thanks for the help!

First, note that $x=2v_3-v_2+v_1$. Now,

$A^kx = A^k(2v_3-v_2+v_1) = 2(A^kv_3) - A^kv_2 + A^kv_1$.

Use the fact that if $\lambda$ is an eigenvalue of a matrix C with eigenvector $v$ then $\lambda ^k$ is en eigenvalue of $C^k$ with eigenvector $v$ (prove this if you haven't already) to get:

$A^kx = 2(x_3^kv_3) - x_2^kv_2 + x_1^kv_1$

and finish.