Possible eigenvalues

[Tolaso J Kos]

What eigenvalues is possible for a matrix $P$ to have if $P^2=P^T$ holds.

[Demetres]

My previous answer was erroneous as I had assumed that $P$ was a real matrix with real eigenvalues and real eigenvectors. So here we go again:


Since $P^T = P^2$, taking transposes in both sides we get $P = P^TP^T$ and combining with the given equation we get $P = P^4$.

From here it is immediate that if $\lambda$ is an eigenvalue of $P$ then $\lambda = \lambda^4$.

So every eigenvalue is a root of $\lambda(\lambda-1)(\lambda^2 + \lambda + 1)$.

The zero and identity matrices show that $\lambda = 0$ and $\lambda =1$ are possible. The roots of $\lambda^2 + \lambda + 1 = 0$ are also possible. For example, we can take

\[ P = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\]

It is easy to check that $P$ satisfies $P^T = P^2$ and that its eigenvalues are the two roots of $\lambda^2 + \lambda + 1 = 0$.