Help me to Improve my method of creating a diagonal matrix? Based on a past exam question:
Q:Consider the matrix
$$A = \pmatrix{5&-6&0\\4&-5&0\\-1&1&2}$$
with entries from $\mathbb{C}$. Find a diagonal matrix $D$ and an invertible matrix $P$ such that $P^{−1}AP = D$.
 A: My attempt was as follows
Using the characteristic polynomial $\chi_A(\lambda)= 0$. I found the eigenvalues to be $\lambda= 2, -1, 1$. All of multiplicity = $1$.
Next use the formula $(\lambda I_n - A)$ and find the basis of the $U_\lambda$ eigenspace for each $\lambda= 2, -1, 1$.
This is done by performing the calculation $(\lambda I_n - A) = A'$, reducing A' by elemenry row operations, and then finding a vector $U=(u_1,u_2,u_3)$ such that using the $A'*U=0$.
I found this to be:
"the basis of $U_2$ is $\left\lbrace \pmatrix{0\\0\\1} \right\rbrace$"
"the basis of $U_{-1} is \left\lbrace \pmatrix{0\\0\\0} \right\rbrace$"
"the basis of $U_{1} is \left\lbrace \pmatrix{-1/2\\1/2\\1} \right\rbrace$"
Since unless stated otherwise, we are working with the standard basis, hence the transition matrix is
$$P = \pmatrix{0&0&1\\0&0&0\\-1/2&1/2&1}$$
$P^-1$ can be found by the forumal $${1/det(P)} * Cof(P)^T$$
Where the $det(P)$ = the determinant of $p$, and $cof(P)^T$ is the transpose of the cofactor matrix
However I found the $det(P)=0$ meaning by the formula, $0=D$, and the diagonal matrix does not exist?

I assume my method to find Eigenvectors are at fault, I ask for advise on how to properly find the basis of the eigenspaces or the eigenvectors, and what the true answer is.
A: the eigenvector is never (0,0,0)...
$U_{-1}\cdots\\
(A-(-1)I)U_{-1} =0\\ \begin{bmatrix}6&-6&0\\4&-4&0\\-1&1&3\end{bmatrix}U_{-1}=0\\
$
$U_{-1}=\begin{bmatrix}1\\1\\0\end{bmatrix}$
$U_1\cdots\\
(A-I)U_1 =0\\ \begin{bmatrix}4&-6&0\\4&-6&0\\-1&1&1\end{bmatrix}U_1=0\\
$
$U_1=\begin{bmatrix}3\\2\\1\end{bmatrix}$
A: Check your working, it is very easy to make a mistake with diagonal matrices, you never have a zero eigenvector 
