If $A$ is symmetric and $B$ is skew-symmetric, then
$tr(AB)=0$.
I would like to know if the converse holds true. If not, can you give me an example?
[Math] trace 0 implies product of symmetric and skew symmetric
analysislinear algebra
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Best Answer
Let $C$ be any real nonsingular $3\times3$ matrix with a zero trace. Then $C$ is not a product of a square matrix (symmetric or not) and a skew-symmetric matrix, because every $3\times3$ skew-symmetric matrix is singular.