# Square of antisymmetric matrix is symmetric and negative definite

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!

• The first one should not be that difficult. Just play around with the definition of an anti-symmetric / symmetric matrix and the properties of transpose. – thanasissdr May 7 '18 at 12:32
• Regarding the second one, the eigenvalues of a real skew-symmetric matrix are all imaginary. Taking this as a fact and the fact that if $\lambda$ is an eigenvalue of $M$, then $\lambda^2$ is an eigenvalue of $M^2$, we can prove the second statement. – thanasissdr May 7 '18 at 12:39

$$(M^2)^T=(M^T)^2=(-M)^2=M^2$$ 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.
• @user107224 $det(M^2)=det(M)^2$ hence $M$ is definite if and only if $det(M) \ne 0$ – stity May 7 '18 at 16:23