Q:

Permutation Matrix

I need to find a Permutation matrix, for an E matrix, i have to permute it because i need to use always two of the rest of eigenvalues of matrix A to operate with them, matrix A is numerically defined by: 
\begin{equation} A=\begin{pmatrix} 1.0000  &   1.0000  &  0.0100 &   0.4900\\
         0   & 1.0000  &  0.0100  &  0.9900\\
         0   & 0       &  0       &  1.0000\\
   -1.0000   &-1.5000   &     0    &  0
\end{pmatrix}\end{equation}
Its eigenvectors matrix E is defined by:
\begin{equation} E=\begin{pmatrix}  0.7909 + 1.3017i   &0 & 0 & 0 \\
         0   & 0.7909 - 1.3017i   &  0       &  0\\
         0   & 0                  &  0.4331  &  0\\
         0   & 0                  &     0    &  -0.0149
\end{pmatrix}\end{equation}
The permutation matrix that i have been using is:
\begin{equation} P=\begin{pmatrix}  0  &0 & 0 & 1 \\
         1   & 0   &  0 &  0\\
         0   & 1   &  0 &  0\\
         0   & 0   &  1 & 0
\end{pmatrix}\end{equation}
To permute matrix E i use the following equation:
\begin{equation} E_p=P^{-k}EP^{k}\end{equation}
where $k=6$ that lets to obtain the different permutations, in this case k would go from 1 to 6 to obtain 6 permutations. This matrix, puts two of the 4 eigenvalues in the first two columns, it doesn't matter the position of the two eigenvalues, that is what i need, but there are two cases for the pair of eigenvalues $\lambda_1,\lambda4=0.7909+1.3017i,-0.0149$ and $\lambda_2,\lambda4=0.7909-1.3017i,-0.0149$; where it doesn't happen because for permutation k=5 and k=6 it is repeated as cases k=1 and k=2. 
Hope you can help me! Thanks a lot.

A:

There is no $4\times4$ permutation matrix with $6$ distinct powers --- there is no element in the symmetric group $S_4$ on four letters with order $6$. The best you can do is order $4$. You need to go to $5\times5$ to get order $6$. 

