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 ?

The problem is that you rounded. The 0.009 in the second row is in fact a 0.
