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?

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

  • $\begingroup$ Wait why doesn't case 1 disprove the anti-symmetric case as well? (including 2x2) $\endgroup$ – Double AA Oct 23 '14 at 4:41
  • $\begingroup$ 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. $\endgroup$ – Cameron Williams Oct 23 '14 at 4:42
  • $\begingroup$ 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. $\endgroup$ – Double AA Oct 23 '14 at 6:19

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