I asked

this question in the homework forum, but desided to move it here, since I'm not in a hurry.

My problem:

How do I find the eigenvalues of this matrix:

$\displaystyle

$$\displaystyle \begin{bmatrix}1&-3&2&-1\\-3&9&-6&3\\2&-6&4&-2\\-1&3&-2&1\end{bmatrix}

$

(Notice that it is symmetric, if that helps)

As I understand I need to solve $\displaystyle {\lambda} $ from:

det

$\displaystyle

\begin{bmatrix}1-{\lambda}&-3&2&-1\\-3&9-{\lambda}&-6&3\\2&-6&4-{\lambda}&-2\\-1&3&-2&1-{\lambda}\end{bmatrix}

$ = 0

The only way I know how to do this is by cofactor expansion. But that would take all day for me. There's got to be a faster way.

galactus said:

" Your matrix should be:

$\displaystyle \begin{bmatrix}{\lambda}-1&3&-2&1\\3&{\lambda}-9&6&-3\\-2&6&{\lambda}-4&2\\1&-3&2&{\lambda}-1\end{bmatrix}$

This gives a characteristic polynomial of:

$\displaystyle {\lambda}^{4}-15{\lambda}^{3}={\lambda}^{3}({\lambda}-15)=0$

"

Why is galactus using a different matrix, and how did he get it? And how did he get the characteristic polynomial of that matrix?

I've been working on this problem for several days now. I just can't solve it. I would be so happy if anyone could help me.

/Gordina