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.


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.

  • $\begingroup$ 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

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