Square root of minus identity

Question

I want to prove that $-I_n$ (seen as real matrix) has a square root if and only if $n$ is even (where $A$ has a square root if it exists $B$ such that $A=B^2$).
If $n$ is even I have found $B$ matrix with diagonal block:
$$\begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix}$$ and this works! For the converse, however, how could I proceed?

Answer 1

In the real case, if $A^2 = - I_n$ with $n$ odd, taking the determinant gives $$\mathrm{det}(A)^2 = -1$$ which is impossible.