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.
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 $6$.