# Find an invertible matrix for this equality

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.
• 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 $A^2 \neq 0$ May 1 '20 at 16:01