I have to solve the following linear algebra problem:
Consider the vector space formed by the set $M_{n}(\mathbb{R})$ of all real square matrices of order $n$ with the usual operations. A scalar product $\langle\cdot,\cdot\rangle:M_{n}(\mathbb{R})\times M_{n}(\mathbb{R}) \to \mathbb{R}$ can be defined as $\langle A,B\rangle=\mathrm{tr}(BA)$. Is this scalar product degenerate?
I "know" this product is non-degenerate because I did solve the case $n=2$ and I guess that it's valid for all $n\in\mathbb{N}$. However, I have no clue of how to solve the general case. I did note that, given two matrices $A=(a_{ij})$ and $B=(b_{ij})$, it follows that $$ \langle A,B \rangle = \mathrm{tr}(BA) = \sum_{i=1}^n B_{i}\cdot A^{i},$$ where $B_{i}$ is the $i$-th row of the matrix $B$ and $A^{i}$ is the $i$-th column of the matrix $A$, so I thought that I could use the non-degeneracy of the dot product and extend it to the matrix scalar product, but so far I haven´t been able to do that. Any hint on how to solve this problem? (I don't want a solution, just a hint).
Hint. By definition, the scalar product is degenerate if there exists some nonzero $A$ such that $\langle A,B\rangle=0$ for every $B$. This, in particular, implies that $\langle A,B\rangle=0$ for every rank-one matrix $B$.