Solution
= Solution
{c}
Let $E_{ji}$ be the <matrix unit> having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\operatorname{tr}(AE_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. Thus
$$
\boxed{R=\{0\}.}
$$
The <matrix trace pairing> is therefore a <nondegenerate bilinear form>, despite its failure to be positive definite when $n\geq2$.