Diagonalize matrix I am sorry if this is stupid in advance. I know how to find eigenvalues of a matrix, but I don't understand how I am suppose to know in which order I am supposed to apply them, in order to find eigenvectors?
Example:
$$
        A=\begin{pmatrix}
        -3 & 2 & 2 \\
        2 & -3 & 2 \\
        2 & 2 & -3 \\
        \end{pmatrix}
$$
Which has eigenvalues of -5,-5,1.
Eigenvector of -5 is:
$$
        E(-5)=y\begin{pmatrix}
        -1\\
        1 \\
        0 \\
        \end{pmatrix}+z\begin{pmatrix}
        -1\\
        0 \\
        1 \\
        \end{pmatrix}
$$
Eigenvector of 1 is
$$
        E(1)=y\begin{pmatrix}
        1\\
        1 \\
        1 \\
        \end{pmatrix}
$$
So basically that's ok. Now, how I am supposed in which order I should reorganize these eigenvectors? I mean, the P could be:
$$
        P_1=\begin{pmatrix}
        -1 & -1 & 1 \\
        1 & 0 & 1 \\
        0 & 1 & 1 \\
        \end{pmatrix}=E(-5)+E(1)
$$
But it also could be 
$$
        P_2=\begin{pmatrix}
        1 & -1 & -1 \\
        1 & 1 & 0 \\
        1 & 0 & 1 \\
        \end{pmatrix}=E(1)+E(-5)
$$
But the $P_1^{-1}$ and $P_2^{-1}$ are different! Thus, $P_2^{-1}AP$ doesn't give an diagonalized matrix, but $P_1^{-1}AP$ does. 
What did I miss? How I can be sure to rearrange eigenvector in right order before proceeding to calculate $P^{-1}$?
EDIT Ok thank you very much, I should have made some mistake when calculating one of inverse matrices. I understood that rearranging eigen(values/vectors) doesn't change the fact that I will still must get a diagonalized matrix at the end.
 A: $P_1^{-1} A P_1 = \begin{bmatrix} -5\\&-5\\&&1 \end{bmatrix}\\
P_2^{-1} A P_2 = \begin{bmatrix} 1\\&-5\\&&-5 \end{bmatrix}$
Here is what I think about. 
$A\mathbf v_1 =  \lambda_1\mathbf v_1\\
A\begin{bmatrix} \mathbf v_1,\mathbf v_2,\mathbf v_3\end{bmatrix} =  \begin{bmatrix} \mathbf v_1,\mathbf v_2,\mathbf v_3\end{bmatrix}\begin{bmatrix} \lambda_1\\&\lambda_2\\&&\lambda_3\end{bmatrix}$
A: $P_1^{-1}AP_1$ are $P_2^{-1}AP_2$ are both diagonal matrices.
You are just reordering the eigenvalues for the diagonal matrices.
$$P_1^{-1}AP_1 = diag(-5, -5, 1) $$
$$P_2^{-1}AP_2 = diag(1, -5, -5) $$
If $Av_j = \lambda_j v_j$, then 
$$A \begin{bmatrix} v_1 & \ldots & v_n\end{bmatrix} = \begin{bmatrix} v_1 & \ldots & v_n\end{bmatrix}diag(\lambda_1 , \ldots , \lambda_n) $$
Let $P = \begin{bmatrix} v_1 & \ldots & v_n\end{bmatrix}$ and $D= diag( \lambda_1 , \ldots , \lambda_n)$,
then we have $$AP=PD$$ and hence
$$P^{-1}AP=D$$
Note that the order of $v_i$ controls the order of $\lambda_i$ in the diagonal matrix.
