Solution
= Solution
The <cyclic property of the trace> gives $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, which is C1. The <linearity> of <trace> and <matrix multiplication> gives C2 and C3. Finally symmetry of $A$ gives
$$
\langle A,A\rangle=\operatorname{tr}(A^2)=\sum_{i,j}A_{ij}A_{ji}=\sum_{i,j}A_{ij}^2\ge0,
$$
with equality exactly when every entry is zero. Thus C4 also holds. This is the <Frobenius inner product> restricted to the real <vector space> of <symmetric matrices>.