# eigenvectors

• May 6th 2008, 09:35 AM
alireza1989
eigenvectors
Hi,
I have a probelm in finding eigenvectors. For example, we have a matrix:
1 2
3 2

The eignevalues for this are x = -1 and x = 4. I understand how to find the eigenvector when x = -1. But when x = 4, we get:
Code:

-3    2      y    0         *      =  3    -2      z    0
which, when multiplied by means that 3y - 2z = 0 (using (A-xI)v=0). Now the final answer is:
Code:

        2y v =           3y
How??
• May 6th 2008, 02:05 PM
topsquark
Quote:

Originally Posted by alireza1989
Hi,
I have a probelm in finding eigenvectors. For example, we have a matrix:
1 2
3 2

The eignevalues for this are x = -1 and x = 4. I understand how to find the eigenvector when x = -1. But when x = 4, we get:
Code:

-3    2      y    0         *      =  3    -2      z    0
which, when multiplied by means that 3y - 2z = 0 (using (A-xI)v=0). Now the final answer is:
Code:

        2y v =           3y
How??

What are you doing with that last matrix?

$\left ( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right ) \cdot \left ( \begin{matrix} y \\ z \end{matrix} \right ) = \lambda \left ( \begin{matrix} y \\ z \end{matrix} \right )$
$\left ( \begin{matrix} 1 - \lambda & 2 \\ 3 & 2 - \lambda \end{matrix} \right )$