I am trying to prove that if $-a^2,-b^2$ are distinct non-zero eigenvalues of $M^2$ where $M$ is a real square skew-symmetric matrix, then there exist orthogonal vectors $\mathbf{v},\mathbf{v}',\mathbf{w},\mathbf{w}'$ such that:

$$M\mathbf{v}=a\mathbf{v}'\qquad M\mathbf{w}=b\mathbf{w}'$$ $$M\mathbf{v}'=-a\mathbf{v}\qquad M\mathbf{w}'=-b\mathbf{w}$$

My ideas:

$M$ has purely imaginary eigenvalues. Let $\mathbf{u}$ be an eigenvector of $M$ and let $M\mathbf{u}=ia\mathbf{u}$ for $a$ real. Then $M^2\mathbf{u}=-a^2\mathbf{u}$. In the same direction, $M\mathbf{r}=ib\mathbf{r}$ gives $M^2\mathbf{r}=-b^2\mathbf{r}$. Since $M^2$ is symmetric we know that $\mathbf{u},\mathbf{r}$ are orthogonal. So I thought setting $\mathbf{v}=\mathbf{u},\;\mathbf{v}'=i\mathbf{u},\;\mathbf{w}=\mathbf{r},\;\mathbf{w}'=i\mathbf{r}$ would work (this satisfies the conditions above), but the problem is that $\mathbf{v},\mathbf{v}'$ are not orthogonal (and $\mathbf{w},\mathbf{w}'$ are not either). Am I on the right track at least?


I think that you should start from the eigenvectors of $M^2$ to be sure you find real vectors (the question implicitly asks for real things). So take $v$ and $w$ as eigenvectors of $M^2$ with respective eigenvalues $-a^2,-b^2$ and take $v'$ and $w'$ given by the condition $$M\mathbf{v}=a\mathbf{v}'\qquad M\mathbf{w}=b\mathbf{w}' $$

Then you can prove that $v$ and $v'$ are orthogonal. You should then also prove that $v$ and $w'$ are orthogonal which comes from the fact that also $v'$ and $w'$ turn out to be eigenvectors of $M^2$.

  • $\begingroup$ Thanks. In the first step, how can I prove the existence of such vectors? $\endgroup$ – WPL Jan 7 '14 at 14:06
  • $\begingroup$ The matrix should be $M^t\cdot M$ not $M^2$. Now that matrix is symmetric... $\endgroup$ – Emanuele Paolini Jan 7 '14 at 14:11
  • $\begingroup$ Uhm... actually $M^2$ is also symmetric... I'm not sure if you should use $M^2$ or $M^t M$. I haven't checked all the details. $\endgroup$ – Emanuele Paolini Jan 7 '14 at 14:13
  • $\begingroup$ Yes $M^2$ is symmetric. But am I aloud to just say "take $v,v',w,w'$" such that "..." without justification? The problem is that I can't see why this must be true. :S $\endgroup$ – WPL Jan 7 '14 at 14:20
  • $\begingroup$ I'm not sure I understand what is the problem... anyway I made some small modification to the answer. $\endgroup$ – Emanuele Paolini Jan 7 '14 at 14:45

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