Square of antisymmetric matrix is symmetric and negative definite

Question

Prove that if $M$ is antisymmetric, then $M^2$ is

1. symmetric, and

2. negative definite.

I have no idea where to begin. I tried to diagonalise the matrix and attempt something with $D^2 = S^T M^2 S$, but I don’t really know what I can do with this information, especially since I don’t have information about the eigenvalues of $M$. Could someone help me proceed from here? Thanks!

Answer 1

$$(M^2)^T=(M^T)^2=(-M)^2=M^2$$ hence $M^2$ is symmetric

Let $x\in \mathbb{R}^n$, $$(x,M^2x)=x^TM^2x=x^T(-M^T)Mx=-x^TM^TMx=-\|Mx\|_2^2 \le0$$ hence $M^2$ is negative semidefinite.

You need additional conditions to prove that $M^2$ is negative definite. As stated, $M$ could, for example, be the null matrix.