Please correct my work, finding Eigenvector Determine whether matrix A is a Diagonalizable. if it is , determine matrix P that Diagnolizes it and compute $P^{-1}AP$. 
$$A=
        \begin{bmatrix}
        +3 & +2\\
        -2 & -3\\
        \end{bmatrix}
$$
$$A-\ell I = 
        \begin{bmatrix}
        +3-\ell & +2\\
        -2 & -3-\ell\\
        \end{bmatrix}
$$
Then Determinant should be zero :
$$
        \begin{vmatrix}
        +3-\ell & +2\\
        -2 & -3-\ell\\
        \end{vmatrix}=(3-\ell)(-3-\ell)+4=0 \to \ell^2-5=0 \to  \ell_{1}=\sqrt 5,\ell_{2}=-\sqrt5 ,
$$
$$
        \begin{bmatrix}
         3-\ell_{1}& +2\\
        -2 & -3-\ell{1}\\
        \end{bmatrix}=    
        \begin{bmatrix}
         3-\sqrt 5=0.76& +2\\
        -2 & -3-\sqrt 5=-5.24\\
        \end{bmatrix}\to{R_1\Leftarrow\Rightarrow R2 \mapsto} 
        \begin{bmatrix}
        -2 & -5.24\\
         0.76& +2\\
        \end{bmatrix}\to{R_1=R1/{-2}\mapsto} 
        \begin{bmatrix}
        1 & 2.62\\
        0.76& +2\\
        \end{bmatrix}\to{R_2=R2-0.76R1\mapsto} 
        \begin{bmatrix}
        1 & 2.62\\
        0 & 0.009\\
        \end{bmatrix}\to{R_2=R2/0.009\mapsto} 
        \begin{bmatrix}
        1 & 2.62\\
        0 & 1\\
        \end{bmatrix}\to{R_1=R1-2.62R1\mapsto} 
        \begin{bmatrix}
        1 & 0 | 0\\
        0 & 1 | 0\\
        \end{bmatrix}
$$
$then$
$$ V_1=       \begin{bmatrix}
        0 \\
        0 \\
        \end{bmatrix}
$$
I am stuck here , matrix of $0,0$ is the right answer Eigenvector 1 ?
 A: The eigenvalues are correct. However you go wrong in the computation of the eigenvectors:
$$
\begin{bmatrix}
3-\sqrt{5} & 2\\
-2 & -3-\sqrt{5}
\end{bmatrix}
$$
Divide the first row by $3-\sqrt{5}$, which is the same as multiplying it by $(3+\sqrt{5})/2$, getting
$$
\begin{bmatrix}
1 & \frac{3+\sqrt{5}}{2}\\
-2 & -3-\sqrt{5}
\end{bmatrix}
$$
Now adding to the second row the first one multiplied by $2$ brings the matrix in the form
$$
\begin{bmatrix}
1 & \frac{3+\sqrt{5}}{2}\\
0 & 0
\end{bmatrix}
$$
so you know that one eigenvector is
$$
\begin{bmatrix}
-\frac{3+\sqrt{5}}{2}\\
1
\end{bmatrix}
$$

The computations for the other eigenvalue are similar
$$
\begin{bmatrix}
3+\sqrt{5} & 2\\
-2 & -3+\sqrt{5}
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & \frac{3-\sqrt{5}}{2}\\
-2 & -3+\sqrt{5}
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & \frac{3-\sqrt{5}}{2}\\
0 & 0
\end{bmatrix}
$$
So the other eigenvector you're looking for is
$$
\begin{bmatrix}
-\frac{3-\sqrt{5}}{2}\\
1
\end{bmatrix}
$$

Of course a matrix that diagonalizes $A$ is
$$
P=
\begin{bmatrix}
-\frac{3+\sqrt{5}}{2} & -\frac{3-\sqrt{5}}{2}\\
1&1
\end{bmatrix}
$$
A: The problem is that you rounded. The 0.009 in the second row is in fact a 0.
