Express a power of an orthogonal matrix in different ways

Question

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 with Mathematica 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:

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

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

Let's make the eigen decomposition of $M^2$ and check for equalities:

dp and Pp mean $d'$ and $p'$

$\{\text{dp},\text{Pp}\}=\text{Eigensystem}[M.M]$

$\text{Pp}=\text{Pp}^T$

$\text{dp}=\text{DiagonalMatrix}[\text{dp}]$

$Round[Pp.dp.Inverse[Pp]] == M.M$ (TRUE)

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

$d.d == dp$

$P == Pp$

Summarizing:

$P\cdot d^2 \cdot P^{-1} = M^2$

$P' \cdot d'^2 \cdot P'^{-1} = M^2$

$d^2 \neq d'$

$P \neq P'$

We have seen that Eigenvectors of $M$ and $M^2$ are different and Eigenvalues of $M^2$ are not $\lambda^2$, but yet you can write $M^2$ in two different expressions.

Why $M^2$ can be written in these two previous expressions?

Answer 1

The test DiagonalMatrix[Eigensystem[M.M][[1]]] == d.d is flawed. There is no reason to expect that the order in which Mathematica returns the eigenvalues of $M$ will exactly match the order of the corresponding eigenvalues of $MM$. If you examine the two lists yourself, you’ll see that the eigenvalues of $MM$ are indeed equal to the squares of the eigenvalues of $M$, but that the two lists are ordered differently. Instead of the above test, you need something along the lines of Sort[Eigenvalues[M.M]] == Sort[Eigenvalues[M]^2]. (N.B.: there’s no need to expand the list of eigenvalues into a diagonal matrix in order to square them since the ^ operator is automatically threaded over lists.)

As for your larger question, the eigendecomposition of a matrix is in general not unique. The eigenvalues can be taken in any order, for one thing. Even if you establish some canonical order for them and insist on unit vectors in the eigenbasis, unless all of the eigenspaces are one-dimensional, any basis of the individual eigenspaces will do, while if they are all one-dimensional there’s still a choice of sign to be made.