Are anticommuting involutory matrices Hermitian?

Question

If $\alpha$ and $\beta$ are matrices such that $\alpha^2=\beta^2=I$, and they anticommute, i.e $\{\alpha,\beta\}=0$. Then, does this imply that $\alpha$ and $\beta $ are Hermitian?

Answer 1

No, here's a counterexample: $$ \alpha = \begin{pmatrix} -1 & \sqrt{2} \\ 0 & 1 \end{pmatrix} \qquad \beta = \begin{pmatrix} 1 & 0 \\ \sqrt{2} & -1 \end{pmatrix} $$

Answer 2

Since $\alpha^2-I=0$, there is a change of basis $M$ such that $M\alpha M^{-1}=\begin{bmatrix}I_k&0\\ 0& -I_h\end{bmatrix}$. Now, let $M\beta M^{-1}=\begin{bmatrix}A&B\\ C& D\end{bmatrix}$. $$\alpha\beta+\beta\alpha=\begin{bmatrix}2A& 0\\ 0& -2D\end{bmatrix}$$ therefore $M\beta M^{-1}=\begin{bmatrix}0&B\\ C& 0\end{bmatrix}$. However $I_n=(M\beta M^{-1})^2=\begin{bmatrix}BC& 0\\ 0 & CB\end{bmatrix}$, therefore $k=h=\frac n2$ and $C=B^{-1}$.

Vice versa, if $n$ is even and there is a change of basis that puts $\alpha$ in the form $\begin{bmatrix}I_{n/2}&0\\ 0& -I_{n/2}\end{bmatrix}$ and $\beta$ in the form $\begin{bmatrix}0& A\\ A^{-1}& 0\end{bmatrix}$, then the matrices have those properties of yours.

Putting all together, the following are equivalent:

1. $\alpha^2=\beta^2=I_n$ and $\alpha\beta+\beta\alpha=0$.
2. $n$ is even and there are an invertible matrix $M$ and an $\frac n2\times \frac n2$ matrix $A$ such that $M\alpha M^{-1}=\begin{bmatrix}I_{n/2}&0\\ 0& -I_{n/2}\end{bmatrix}$ and $M\beta M^{-1}=\begin{bmatrix}0&A\\ A^{-1}& 0\end{bmatrix}$.
3. $n$ is even, there is an invertible matrix $M$ such that $M\alpha M^{-1}=\begin{bmatrix}I_{n/2}&0\\ 0& -I_{n/2}\end{bmatrix}$ and, for all matrices $H$ such that $H\alpha H^{-1}=\begin{bmatrix}I_{n/2}&0\\ 0& -I_{n/2}\end{bmatrix}$, there is an $\frac n2\times \frac n2$ matrix $A$ such that $H\beta H^{-1}=\begin{bmatrix}0&A\\ A^{-1}& 0\end{bmatrix}$.

Now it is clear that $\begin{bmatrix}0& A\\ A^{-1}& 0\end{bmatrix}$ isn't Hermitian unless $A$ is unitary.

For an instance where neither is, consider $$\alpha=\begin{bmatrix}1&5\\ 2&1\end{bmatrix}^{-1}\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}1&5\\ 2&1\end{bmatrix}\\\beta=\begin{bmatrix}1&5\\ 2&1\end{bmatrix}^{-1}\begin{bmatrix}0&1\\ 1&0\end{bmatrix}\begin{bmatrix}1&5\\ 2&1\end{bmatrix}$$

and all that can be done with these.