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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!

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    $\begingroup$ 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. $\endgroup$ – thanasissdr May 7 '18 at 12:32
  • $\begingroup$ 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. $\endgroup$ – thanasissdr May 7 '18 at 12:39
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$$(M^2)^T=(M^T)^2=(-M)^2=M^2$$ 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.

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  • $\begingroup$ Hi there, what kind of constraints should I be looking to place on M for negative definite M? Should I be looking more at the determinant? $\endgroup$ – user107224 May 7 '18 at 13:35
  • $\begingroup$ @user107224 $det(M^2)=det(M)^2$ hence $M$ is definite if and only if $det(M) \ne 0$ $\endgroup$ – stity May 7 '18 at 16:23

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