$tr(AB) = 0$ for all $B \in M_n(\mathbb{R})$ implies $A = 0$?

Question

Let $V= M_n(\mathbb{R})$. I'm trying to prove that the bilinear form $$g:V \times V \rightarrow \mathbb{R}$$ $$g(A,B) = tr(AB),$$ is non-degenerate. To prove it, I tried to show that the linear transformation $T:V \rightarrow V^*$, $T(A)(B) = g(A,B)$ is an isomorphism. But, I didn't make any progress.

Answer 1

If all you want is to show $g$ is non-degenerate, that's definitely possible. Suppose $A\in M_{n\times n}(\Bbb{R})$ is such that $g(A,\cdot)=0$. Then, in particular, $g(A,A^t)=\text{tr}(AA^t)=0$. Now, expanding out what the trace is, you can convince yourself it is $\sum\limits_{i,j=1}^na_{ij}^2$. If a sum of squares of real numbers is $0$, then each summand is zero, and hence each $a_{ij}=0$; i.e $A=0$. This proves non-degeneracy.

Answer 2

Let $$ \delta_{(i,j)}(k,\ell)=\begin{cases} 1, & \mbox{ if } (k,\ell)=(i,j)\\ 0, & \mbox{ if } (k,\ell)\neq (i,j)\\ \end{cases} $$ and the $n\times n$ matrix $E^{k,\ell}=\Big(\delta_{(i,j)}(k,\ell)\Big)_{n\times n}$. Fix $A=\big( a_{ij}\big)_{n\times n}$. Note that for all $E^{k,\ell}$ we have $$ \mathrm{trace}\Big( A\cdot E^{k,\ell} \Big)=a_{k \ell}=0. $$ So what? What can you conclude?