I have matrix $\displaystyle A=\begin{pmatrix}0&1&0\\2&-2&-1\\-3&0&0\end{pmatrix}$

and I get that $\displaystyle det(A-LE)=-L^3+2L^2+2L+3$

so

$\displaystyle L^3-2L^2-2L-3=0$

//after this our lecturer takes random numbers and checks if they can be eigenvalues.

For this problem I'm sure that eigenvalue is only $\displaystyle 3$

//but what if I get $\displaystyle L^3-7L^2-36=0$?

//I have to check all numbers till 6

//so how did he find that for this one $\displaystyle L$ is $\displaystyle 6, 3$ and $\displaystyle 2$ so fast?

//EH... NEVERMIND xD

further, to find eigenvector I have

$\displaystyle \begin{pmatrix}-3&1&0&|0\\2&-5&-1&|0\\-3&0&-3&|0\end{pmatrix}$

but I can't find a way to substitute so that I would have 2 even rows or columns, so maybe there is another way I don't know?

some update:

I ended up at

$\displaystyle \begin{pmatrix}-3&2&-1&|0\\0&-3&-3&|0\end{pmatrix}=\begin{pmatrix}1&-2/3&1/3&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&-0&1&|0\\0&1&1&|0\end{pmatrix}=\begin{pmatrix}1&0 &|-C\\0&1&|-C\end{pmatrix}$

$\displaystyle x_1 = -C$

$\displaystyle x_2 = -C$

$\displaystyle x_3 = C$

So, after getting someone annoying for updating my post so many times xD

Q: HOW DO I STANDARDIZE EIGENVECTOR AND CHECK IF ANSWER IS CORRECT?

someone told me: "you need to find vector length or smth" =/

$\displaystyle |x|=\sqrt{(-C)^2+(-C)^2+C^2}=\sqrt{3}C$

$\displaystyle x_N=\begin{pmatrix}-1/\sqrt{3}\\-1/\sqrt{3}\\1/\sqrt{3}\end{pmatrix}$ ???????

IF CORRECT, HOW DO I CHECK IF IT IS SO?