# Eigenvectors

• Jul 1st 2008, 07:00 PM
jschlarb
Eigenvectors
I've calculated the following eigen value and determined the following matrix:

eigenvalue, U= -5

Original matrix
([-5-U] 2 -3)
(0 [1-U] -2)
(0 1 [4-U])

Solved Matrix:

(0 0 28)0
(0 3 -1)0
(0 0 0) 0

I'm not exactly clear on how to determined the eigenvector
• Jul 1st 2008, 07:31 PM
mr fantastic
Quote:

Originally Posted by jschlarb
I've calculated the following eigen value and determined the following matrix:

eigenvalue, U= -5

Original matrix
([-5-U] 2 -3)
(0 [1-U] -2)
(0 1 [4-U])

Solved Matrix:

(0 0 28)0
(0 3 -1)0
(0 0 0) 0

I'm not exactly clear on how to determined the eigenvector

Solve MX = -5X for X, where M is the given matrix and X = [x, y, z]^T is the eigen vector.

You should get x = t, y = z = 0. So the eigenvector corresponding to the eigenvalue u = -5 is [1, 0, 0]^T.
• Jul 1st 2008, 09:02 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Solve MX = -5X for X, where M is the given matrix and X = [x, y, z]^T is the eigen vector.

I should have added that there are two other eigenvalues: u = 2 and u = 3 ......