TITLE: Does there exist a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$ such that $\mathbf{A}^{2}=-\mathbf{I}$?
QUESTION [15 upvotes]: Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,
$$\mathbf{A}^2=-\mathbf{I}$$
I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?

REPLY [3 votes]: Here is an argument without using determinant or eigenvalues (yes, this is a deliberately complicated answer). We prove by contradiction. Suppose $A^2=-I_3$. Take $x\neq0$ and let $y=Ax$. If $y=kx$ for some $k\in\mathbb{R}$, then
$$k^2x=ky=kAx=A(kx)=Ay=A^2x=-x\ \Rightarrow\ k^2=-1,$$
which is impossible. Therefore $x$ and $y$ are linearly independent. Extend them to a basis $\{x,y,z\}$ of $\mathbb{R}^3$. Let $w=Az$. By similar reasoning, $z$ and $w$ are linearly independent and hence $w\neq0$. Therefore
$$w=ax+by+cz\tag{1}$$
for some $(a,b,c)^T\in\mathbb{R}^3\setminus0$. Then $-z=Aw=A(ax+by+cz)=ay-bx+cw$, i.e.
$$cw=bx-ay-z.\tag{2}$$
As $x,y,z$ are linearly independent, $c\neq0$. Therefore $(2)$ implies that
$$w=\frac{b}{c}x-\frac{a}{c}y-\frac{1}{c}z.\tag{3}$$
Comparing the coefficients of $z$ in $(1)$ and $(3)$, we get $c=-\frac1c$, which is impossible. Hence there does not exist a real matrix $A$ such that $A^2=-I_3$.