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for $\mathbf{A} \in \mathbb{R}^{n \times n}$ satisfying $\mathbf{A}^2 = -\mathbf{I}_n$, what can be concluded about $n$ and $\text{det(A)}$?

Here's my attempt:

let $\det(A) = k$,

$\det(\mathbf{A}^2) = \det(\mathbf{A}) \det(\mathbf{A}) = k^2 = \det(-\mathbf{I_n}) = -1 \implies k^2 = -1 \implies k = \sqrt{-1} = i$

and hence, $\det(A) = i$. But I'm not sure about the $n$ part. Any help would be highly appreciated

user23569449
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    In fact $\mathrm{det} (-\mathbf{I}_n) = (-1)^n$. So one can conclude that n must be even, and $\mathrm{det}(\mathbf{A}) = 1$ or $-1$. – Mingxin Li Nov 17 '24 at 13:48
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    And, of course, $k^2=a$ never has only one solution so, when $a\neq0.$ Even when $a>0.$ $k^2=4$ doesn't mean $k=\sqrt4=2.$| – Thomas Andrews Nov 17 '24 at 14:10

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Consider $A \in \mathbb{R}^{n \times n}$, satisfying $A^2 = -I_n$. Since $det(A^2) = det(A)^2$ and $det(-I_n) = (-1)^n$, applying our hypothesis, we get that $det(A)^2 = (-1)^n$, meaning $det(A) = \pm \sqrt{(-1)^n}$. The only case in which $det(A) \in \mathbb{R}$ is when $(-1)^n = 1$, which happens if and only if $n$ is even.