TITLE: Proving there is no squared $n\times n$ (n odd) real matrix yielding minus the identity matrix
QUESTION [7 upvotes]: Prove or give a counterexample

There is no $A \in \Bbb R^{3 \times 3}$ such that $A^2 = -\Bbb I_3$

Here's my attempt
I suspect that the statement is true. To prove it, I used contradiction
Assuming $A^2 = -\Bbb I_3$ holds, we take the determinant on both sides to get
$$ \det A^2= (\det A)^2= (-1)^3 \det(\Bbb I_3) \Rightarrow \det A = \sqrt{-1}$$
So the determinant of $A$ is ill-defined over the real numbers, which is a contradiction (as the determinant of square matrices is well-defined over the real numbers).
Do you all agree?

REPLY [3 votes]: Here's another related argument. The characteristic polynomial of $A$ is a degree 3 polynomial with real coefficients, so it's got a root. Hence $A$ has at least one real eigenvalue $\lambda$ with eigenvector $\vec{v}$. But then
$$-\vec{v} = -\mathbb{I}_3\vec{v} = A^2 \vec{v} = \lambda^2 \vec{v}$$
so $\lambda^2 = -1$, which is a contradiction.