TITLE: 3$\times$3 and 2$\times$2 matrices that satisfy $A^2 = -I$, $I$ being the identity matrix.
QUESTION [2 upvotes]: I am looking for all square matrices $A$ of order 2 and 3 that satisfy $A^2 = - I$, $I$ being the identity matrix of the corresponding order.

REPLY [2 votes]: @vadmin answers your question for $3\times3$ matrices. For $2\times2$ matrices, either:

There is at least one real eigenvalue. So there is a subspace of dimension at least $1$, where $A$ acts as scaling by $\lambda$. So $A^2$ acts as scaling by $\lambda^2\geq0$. This is not compatible with $A^2$ being $-I$, since $-I$ scales by $-1$ in all directions.
Two complex conjugate eigenvalues: first you can factor out some $cI$ so that $A=(cI)B$ with $\left\lvert B\right\rvert=1$. Now $B$ has two complex conjugate eigenvalues each with magnitude $1$. There is a basis in which $B$ is simply a rotation matrix by some angle $\theta$. Note that $-I$ is rotation by $\pi$ about the origin. To have $A^2=-I$, you need $c^2B^2$ to be rotation by $\pi$. So you need $c=\pm1$ and $B$ to be a rotation by $\pi/2$. 

So the solutions to this issue are 
$$\begin{align}
A&=\pm P^{-1}\begin{bmatrix}\cos(\pi/2)&-\sin(\pi/2)\\\sin(\pi/2)&\cos(\pi/2)\end{bmatrix}P\\
&=\pm P^{-1}\begin{bmatrix}0&-1\\1&0\end{bmatrix}P
\end{align}$$
where $P$ can be any invertible matrix. ($P$ is the change of basis matrix from the standard basis to the one where $B$ is a rotation). If you want to spell it out by entries, then let $a,b,c,d$ be any four numbers with $ad-bc\neq0$. Then you have $$\begin{align}
A
&=\pm\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}\frac{\begin{bmatrix}d&-b\\-c&a\end{bmatrix}}{ad-bc}\\
&=\frac{\pm1}{ad-bc}\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}c&-a\\d&-b
\end{bmatrix}\\
&=\frac{\pm1}{ad-bc}\begin{bmatrix}ac+bd&-a^2-b^2\\c^2+d^2&-ac-bd\end{bmatrix}
\end{align}$$
{"set_name": "stack_exchange", "score": 2, "question_id": 1502868}