I have the following orthogonal matrix:

$M_{5,5}=\left(
\begin{array}{ccccc}
 0 & 1 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 0 & 1 \\
 0 & 0 & 1 & 0 & 0 \\
\end{array}
\right)$

My aim here is to diagonalize it and check some interesting properties of powers of $M$. Then I proceed to express $M$ as $M=P.D.P^{-1}$

>$\{d,P\}=\text{Eigensystem}[M]$

>$\text{P = Transpose[P]}$

>$\text{d = DiagonalMatrix[d]}$


>$\text{Round}\left[P.d.P^{-1}\right]==M$

which is **True**


>$\text{Round}\left[P.d.d.P^{-1}\right]==M.M$

which is **True** also.

This means that $M^k = P.D^k.P^{-1}$ so the eigenvalues of $A^k$ are $\lambda^k=\{D^k_{1,1},D^k_{2,2},D^k_{3,3},D^k_{4,4},D^k_{5,5}\}$

but when I compare $\lambda^2$ with the eigenvalues of $M^2$ the output is **False**.

>$\text{DiagonalMatrix}[\text{Eigensystem}[M.M][[1]]]=d.d$

which is **False**

Now I start to think that $M^2$ can be written in 2 different ways:

$M^2 = P.D^2.D^{-1}$ which we have seen it is true.

$M^2 = Q.D'.Q^{-1}$ this means that other eigen decomposition exists in ${M^2}$.

Can someone tell me If I'm wrong? If true why $M^2$ can be written in these two expressions?