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Suppose that $n\geq 2$ and that $A,B\in M_n(\mathbb{R})$ are such that $A^2=-I_n$ and $AB=BA$. Can we infer that $\det{B}\geq0$?

My attempt:
Since $A^2=-I_n$, we have $A^{-1}=-A$ and $\det{A}=\pm1$. Also, we can deduce that: $$\begin{align*} \det{A^{-1}}&=(-1)^n\det{A}\\ \frac{1}{\det{A}}&=(-1)^n\det{A}\\ 1&=(-1)^n(\det{A})^2 \end{align*}$$ This implies that $n$ must be even. However, I am stuck at this point. I tried to use the characteristic polynomial and show that $B$ is diagonalizable over $\mathbb{C}$. But I could not succeed. Any help would be greatly appreciated.

  • @MarianoSuárez-Álvarez this is only true if $n=2$. – Alma Arjuna Dec 30 '23 at 16:57
  • @MarianoSuárez-Álvarez I'm sorry can you please write the proof completely as an answer? I kinda can't understand it. I highly appreciate it. –  Dec 30 '23 at 17:14
  • The usual proof goes like this: by a change of basis we may assume that $A=\pmatrix{0&-I\ I&0}$. Since $A$ and $B$ commute, $B$ must be in the form of $\pmatrix{X&-Y\ Y&X}$ for some real square matrices $X$ and $Y$. In turn, $B$ is similar over $\mathbb C$ to $\pmatrix{X+iY&\ast\ 0&X-iY}$. Hence $\det B=|\det(X+iY)|^2\ge0$. – user1551 Dec 30 '23 at 17:22
  • @user1551 Thanks, but how did you conclude that $A$ is similar to $\begin{bmatrix}0 & -I\I & 0\end{bmatrix}$? –  Dec 30 '23 at 17:38
  • @user1551 Also I couldn't understand how $B$ is similar to $\begin{bmatrix}X+iY & *\0 & X-iY\end{bmatrix}$. Can you explain? –  Dec 30 '23 at 17:40

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