$\operatorname{tr}(AB) = 0$ for (skew-)symmetric matricies

I know if A is symmetric and B is skew-symmetric then $\operatorname{tr}(AB) = 0$. (This follows because $\operatorname{tr}(AB) = -\operatorname{tr}(AB)$)

Is the converse of that true? In other words, does $\operatorname{tr}(AB) = 0$ where is A is symmetric or skew-symmetric imply B is skew-symmetric or symmetric respectively? Why?

• Wait, why is $tr(AB)=-tr(AB)$? – Nishant Oct 23 '14 at 4:22
• @Nishant $A^T = A$, $B^T = -B$ and the trace is cyclic. – Cameron Williams Oct 23 '14 at 4:22
• Oh, right, I get it now. – Nishant Oct 23 '14 at 4:28

The case of $A$ being symmetric fails to imply that $B$ is anti-symmetric in a pretty trivial case (let's ignore the zero matrix since we often ignore $0$ in many arguments like this): $A = I_{n\times n}$ as this only implies $B$ is traceless, nothing more.

The case of $A$ being anti-symmetric fails to imply that $B$ is symmetric in general as well but for slightly less trivial reasons. Note: if $A$ and $B$ are $2\times 2$ matrices, the anti-symmetry of $A$ does imply the symmetry of $B$. (Check it yourself.) To see that it fails in general we then need to consider $3\times 3$ matrices or larger. Let $A$ and $B$ be given as below.

$$A = \left(\begin{array}{rrr} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right)$$

$$B = \left(\begin{array}{rrr} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & 0 \end{array}\right)$$

Then $\text{tr}(AB) = 0$ and $A$ is anti-symmetric but $B$ is not symmetric.

• Wait why doesn't case 1 disprove the anti-symmetric case as well? (including 2x2) – Double AA Oct 23 '14 at 4:41
• Hah. I guess so but I wanted to neglect the zero matrix for the anti-symmetric case because it's a more interesting problem I think. – Cameron Williams Oct 23 '14 at 4:42
• Ok +1, check and thanks. The only B that would serve for every symmetric/skewsymmetric matrix A would have to be a skewsymmetric/symmetric one though. All these counterexamples only work in very specific circumstances. – Double AA Oct 23 '14 at 6:19