The contraction of a symmetric tensor $S_{\mu\nu}$ with another tensor $A^{\mu\nu}$ is $0$, prove that $A^{\mu\nu}$ is antisymmetric

Question

Is well known that if $S_{\mu\nu}$ is a symmetric tensor and $A^{\mu\nu}$ is an antisymmetric tensor: $$S_{\mu\nu}A^{\mu\nu}=0.$$ I want to prove or disprove that if $$S_{\mu\nu}A^{\mu\nu}=0,$$ and $S_{\mu\nu}$ is symmetric then $A^{\mu\nu}$ must be antisymmetric.

Answer 1

I assume that $S_{μν}A^{μν}=0$ for all symmetric $S$.

Since $S_{μν}A^{μν}=0$ for all symmetric $S$, we can choose $S_{11}=1$ and $S_{\mu\nu}=0$ to all remaining entries of $S$. Then it follows from $S_{μν}A^{μν}=0$ that $$\boxed{A^{11}=0}.$$

Now, choose $S_{12}=S_{21}=1$ and $S_{\mu\nu}=0$ to all remaining entries of $S$. Then it follows from $S_{μν}A^{μν}=0$ that $$A^{12}+A^{21}=0\implies\boxed{A^{12}=-A^{21}}.$$

Repeating this procedure to all remaining entries of $S$, we show that $$\boxed{A^{\mu\nu}=-A^{\nu\mu}}.$$

This means that $A$ is antisymmetric.

Answer 2

Take $S_{\mu\nu} = v_\mu w_\nu + w_\mu v_\nu$, where the numerical values $v_\mu$ and $w_\nu$ are arbitrarily given. Then $$S_{\mu\nu}A^{\mu\nu} = 0$$ reads $$A^{\mu\nu}v_\mu w_\nu = -A^{\mu\nu}w_\mu v_\nu,$$and relabeling indices in the right side yields $$A^{\mu\nu}v_\mu w_\nu = -A^{\nu\mu}v_\mu w_\nu.$$The arbitrariety of $v$ and $w$ now implies that $A^{\mu\nu} = -A^{\nu\mu}$.

In more detail: for fixed indices $\alpha$ and $\beta$, let $v_\mu = \delta_{\mu\alpha}$ and $w_\nu = \delta_{\nu\beta}$ to obtain $A^{\alpha\beta} = -A^{\beta\alpha}$, and then relabel $\alpha$ and $\beta$ back to $\mu$ and $\nu$.

(I think some people in physics call this sort of implication the "quotient theorem" for tensors.)