eigenvectors

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• May 6th 2008, 08:35 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, 01:05 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?

$\displaystyle \left ( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right ) \cdot \left ( \begin{matrix} y \\ z \end{matrix} \right ) = \lambda \left ( \begin{matrix} y \\ z \end{matrix} \right )$
for both of your eigenvalues.

You do not use the matrix
$\displaystyle \left ( \begin{matrix} 1 - \lambda & 2 \\ 3 & 2 - \lambda \end{matrix} \right )$
to find your eigenvectors.

-Dan
• May 6th 2008, 03:07 PM
alireza1989
I found out how to do it.
Thanx