Find all e. values and e. vectors of
A = $\left[\begin{matrix}0&1\0&0\end{matrix}\right]$
I got $\lambda =0$ double root
and e. vectors are $\left(\begin{matrix}1\0\end{matrix}\right)$ and $left(\begin{matrix}1\1\end{matrix}\right)$
Please, tell me whether they are correct?

Question

Find all e. values and e. vectors of 
A = $\left[\begin{matrix}0&1\0&0\end{matrix}\right]$
I got $\lambda =0$ double root
and e. vectors are $\left(\begin{matrix}1\0\end{matrix}\right)$ and $left(\begin{matrix}1\1\end{matrix}\right)$
Please, tell me whether they are correct?

loser66 · Answer

@dumbcow

dumbcow · Answer

the 2nd e vector is (1,1) ?

dumbcow · Answer

i get
$$\lambda = 0$$
e vectors
$$\left(\begin{matrix}0 \ 0\end{matrix}\right) , \left(\begin{matrix}1 \ 0\end{matrix}\right)$$

loser66 · Answer

how to get (0,0) ? @dumbcow

dumbcow · Answer

well
$$A \left(\begin{matrix}0 \ 0\end{matrix}\right) = 0  \left(\begin{matrix}0 \ 0\end{matrix}\right) $$
that makes it an eigenvector right?

loser66 · Answer

yes

loser66 · Answer

I can't guess, have to have logic on it
can I take A * (1,0) =(0,0) this + (1,0) is basis of eigen space?

dumbcow · Answer

not sure, i haven't done this in awhile
i forgot what to do with eigen spaces?

loser66 · Answer

when I go on the normal way, it's trivial that (1,0) is a eigenvector, but it is a 2x2 matrix with double root on eigenvalue, I think we still have another eigenvector.

loser66 · Answer

Ok, forget it. I think I overtake the problem. !! Thank you.