Prove there exists a matrix $B$ such that $AB=BA$ and $B^2 = -I$.

Question

Let $A$ be a real $n \times n$ matrix without real eigenvalues.

Prove there exists a real matrix $B$ such that $AB=BA$ and $B^2 = -I$.

I understand that $n$ is even and $A$ is a nonsingular matrix.

Answer 1

It's true. We assume that $n$ is even.

Let $J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. We consider the real Jordan form of $A$. Its Jordan blocks are in the form of this $2r\times 2r$ matrix

$\begin{pmatrix}aI_2+bJ&I_2&0\cdots&0\\0&aI+bJ&I &0\cdots\\\cdots&\cdots&\cdots&\cdots\\0&\cdots&0&aI+bJ\end{pmatrix}$ where $a,b\in\mathbb{R},b\not= 0$.

It suffices to show the existence of $B$ when $A$ is reduced to one Jordan block.

It suffices to choose $B=diag(J_1,\cdots,J_r)$ where $J_k=J$.