Find eigenvectors of $A$ Let $A=\begin{pmatrix} 0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2 \end{pmatrix}$
I find eignevalues $-1,0,1,2$ and $0$ has multiplicity of $2$.
But I only find one eigenvector of $0$ and all the other eigenvalues have eigenvectors, so anyway $A$ has four eigenvectors and diagonalizable.  But I think $0$ must have $2$ eigenvectors and one of eigenvalue can't have eigenvector. What's wrong with me?    
I corrected matrix!!!!!
Then the C.P is $x(x^3-2x^2-x+2)=x(x+1)(x-1)(x-2)$ and the eigenvalues that I got is right.
 A: $$\begin{align} \det(A-xI) &= \det(\begin{pmatrix}-x&0&0&0\\1&-x&0&-2\\0&1&-x&1\\0&0&1&2-x\end{pmatrix})\\ &=-x(-x(-x(2-x)-1)-2) \\&=-x(2x^2-x^3+x-2)\\&=-x(-x^2(x-2)+x-2) \\&=-x(x-2)(-x^2+1)\end{align}$$
so $x=0,x=2,x=1,x=-1$ are eigenvalues
then we must find X such that :
$(\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2\end{pmatrix}-\lambda_iI)X=0$  
$\to$
$\lambda$=0$ :\begin{pmatrix}0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2\end{pmatrix}X=0 \to x_1=2x_4$ and $x_2=-x_4$and $x_3=-2x_4$ our vector can be $(-2x_4,x_4,2x_4,x_4)$as(2,-1,-2,1) 
$x=2:$
$\begin{pmatrix}-2&0&0&0\\1&-2&0&-2\\0&1&-2&1\\0&0&1&2-2\end{pmatrix}X=0$$\to $$x_1=0,x_2=-x_4,x_3=0\,\,$as $\,\,(0,-1,0,1)$
$x=1:$
$\begin{pmatrix}-1&0&0&0\\1&-1&0&-2\\0&1&-1&1\\0&0&1&2-1\end{pmatrix})X=0$ $\to $
$x_1=0,x_2=-2x_4,x_3=-x_4$
as $\,\,(0,-2,-1,1)$
$x=-1:$
$\begin{pmatrix}1&0&0&0\\1&1&0&-2\\0&1&1&1\\0&0&1&2+1\end{pmatrix}X=0$$\to$$x_1=0,x_2=2x_4,x_3=-3x_4\,\,$as $\,\,(0,2,-3,1)$
A: $0$ is not an eigenvalue of multiplicity $2$. This would give you $5$ eigenvalues for a $4 \times 4$ matrix, which is not possible. So you cannot find more than $1$ eigenvector for the eigenvalue $0$.
