$f(A,B) = trace(A^tMB)$, prove that if f is symmetric then M is symmetric

Question

Let $V$ be the space of $2\times2$ matrices over $\mathbb R$, let $M \in V$, and let $f:V \times V\to \mathbb R$ a bilinear form such that $f(A,B) = trace(A^tMB)$ for all $A,B \in V$. Prove that f symmetric if and only if M is symmetric.

The first "if" is easy:

Assuming $M=M^t$, then $f(A,B)=trace(A^tMB)=trace[(A^tMB)^t]=trace[(MB)^tA]=trace(B^tM^tA)=trace(B^tMA)=f(B,A)$

But I couldn't prove the other direction, any hints?

Answer 1

One way to prove the result is as follows. Note that $$f(A,B)=f(B,A) \implies \operatorname{trace}(A^t MB)-\operatorname{trace}(B^t MA)=0.$$ But $$\operatorname{trace}(B^t MA) = \operatorname{trace}((B^t MA)^t)=\operatorname{trace}(A^t M^t B).$$ Therefore, $$f(A,B)=f(B,A) \implies \operatorname{trace}(A^t MB)-\operatorname{trace}(A^t M^t B)=\operatorname{trace}(A^t (M-M^t)B)=0.$$ Put $B = I$ and $A = M-M^t$ to arrive at $$\operatorname{trace}((M-M^t)^t (M-M^t))=\|M-M^t\|_F^2 = 0 \implies M = M^t.$$

Note that, for any real matrix $X$, $\operatorname{trace}(X^t X)$ is the sum of the squares of its elements.

Answer 2

It suffices to try with $A=\begin{pmatrix}0&1\\0&0 \end{pmatrix}$.

Since $f$ is symmetric: $$ 0 = f(A,I)-f(I,A) =trace(A^\top\,M)-trace(M\,A) =M_{12} -M_{21} $$ Thus $M_{12} =M_{21}$.