0
$\begingroup$

Let $K$ be a field with $char(K) \neq 2$. Let $S$ be an invertible $n \times n$ matrix where $n>1$. Show that there exists an invertible matrix $A$ such that $A^TS+SA=0$.

I am stuck with this problem for a while. If $S$ is the identity matrix then this is just asking for a invertible skew-symmetric matrix, which obviously exists (take the one with only $1$ above and only $-1$ below the diagonal). But for arbitrary $S$ I can't find one.

$\endgroup$
1
$\begingroup$

This is not true. If $A^TS+SA=0$, then $A^T(S+S^T)+(S+S^T)A=A^TS+SA+(A^TS+SA)^T=0$. That is, $(S+S^T)A$ is necessarily skew-symmetric. In particular, $$ S=\pmatrix{1&0&0\\ 0&0&-1\\ 0&1&0},\ A=\pmatrix{a&b&c\\ \ast&\ast&\ast\\ \ast&\ast&\ast}\Rightarrow(S+S^T)A=\pmatrix{2a&2b&2c\\ 0&0&0\\ 0&0&0} $$ is skew-symmetric. This occurs only if $a=b=c=0$, but then $A$ cannot possibly be invertible.

$\endgroup$
2
  • $\begingroup$ thank you, it seems the problem is not correct as stated. Just out of curiosity: Do you think I could repair the statement if instead of asking for an invertible $A$ I just wanted $A^2 \neq 0$ $\endgroup$
    – korn55
    May 1 '20 at 16:01
  • $\begingroup$ @korn55 I don't know. I don't see any obvious reason why the modified statement is true/false. $\endgroup$
    – user1551
    May 1 '20 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.