1. ## 3x3 eigenvectors. Clueless.

I admit this isn't too urgent, so apologies, but I'm posting here because I have exams quite soon and my inability to do this particular problem is getting a bit frustrating, not to mention worrying.

Anyway, I understand how to work out eigenvectors for 2x2 matrices well enough, I think, but have absolutely no idea how to go about it for 3x3 matrices. I've looked around online but I either don't understand the notation used or it's just not dumbed down enough for me, I suppose...

Specifically, anyway, the problem is as follows:

Given the 3x3 matrix:

( 2 -4 0 )
( -4 2 0 )
( 0 0 4 )

I have to work out the eigenvalues and corresponding eigenvectors. I've come up with the eigenvalues 4, 6 and -2, and used Mathematica to check these which are apparently right. But where do I go from here to get the eigenvectors?

I only need help with one of them if someone could just explain the method to me - in especially simple terms note, since I'm apparently incapable of understanding anything more complicated.

Starting with the eigenvalue lamda = 4, for example, and calculating (A - lamda I)X (where A is the original matrix, I is the identity) I've come up with these equations:

-2x_11 - 4x_21 = 0
-4x_11 - 2x_21 = 0

I'm using _ to indicate subscript by the way, not sure if there's a clearer way to do this or not. But anyway these equations seem to contradict each other, so I'm not even sure I've started out properly. I assume the bottom value of the resulting column vector would be 0, but that's as far as I can get.

Any help?

2. Hello,

$\displaystyle A= \begin{pmatrix} 2 & -4 & 0 \\ -4 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}$

The eigenvalues are found the same way as for a 2x2 matrix :

$\displaystyle \chi (\lambda)=det(A-\lambda I_3)$

Hence $\displaystyle \lambda$, eigenvalue, verify :

$\displaystyle \begin{vmatrix} 2-\lambda & -4 & 0 \\ -4 & 2-\lambda & 0 \\ 0 & 0 & 4-\lambda \end{vmatrix} = 0$

Supposing your eigenvalues are correct, now let's look for the eigenvectors.

The property of an eigenvalue is that :

$\displaystyle AX=\lambda X$

So you have to solve, for each eigenvalue :

$\displaystyle \begin{pmatrix} 2 & -4 & 0 \\ -4 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} \lambda x_1 \\ \lambda x_2 \\ \lambda x_3 \end{pmatrix}$

Is this more clear ?
I can try showing you how to do for 4, eg

3. So we have :

$\displaystyle 2x_1-4x_2=4x_1$
$\displaystyle -4x_1+2x_2=4x_2$
$\displaystyle 4x_3=4x_3$

$\displaystyle x_1=-2x_2$
$\displaystyle -4x_1-2x_2=0$
$\displaystyle x_3=x_3$

-> $\displaystyle 8x_2-2x_2=0$

-> $\displaystyle x_2=0$

-> $\displaystyle x_1=0$

So an eigenvector to the eigenvalue 4 will be $\displaystyle \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

For 6, we have :
$\displaystyle \begin{bmatrix} 2x_1-4x_2=6x_1 \\ -4x_1+2x_2=6x_2 \\ 4x_3=6x_3 \rightarrow x_3=0 \end{bmatrix}$

And solve it as to find a generator for x1 and x2...

4. Ahh, that helps a lot. So the eigenvectors are (0, 0, 1), (-1, 1, 0) and (1, 1, 0) I'm pretty sure. My mistake was trying to solve (A - lamda I)X=0 instead of AX = lamda X. Thanks for your help, it's much appreciated!

5. You're welcome