… and a simpler proof for it

A much simpler proof for the more generic algebraic rule on manipulating quadratic forms I posted recently just became apparent to me. All you need to do is to first show that

$${\rm trace}\{\ma{A}^{\rm H} \cdot \ma{B}\} = {\rm vec}\{\ma{A}\}^{\rm H} \cdot {\rm vec}\{\ma{B}\},$$

which is fairly easy because both represent a short-hand notation for $\sum_k\sum_\ell\sum_m a_{\ell,k}^* b_{\ell,m}$. Once you have this rule set you simply break ${\rm trace}\left\{\ma{A} \cdot \ma{X} \cdot \ma{R} \cdot \ma{X}^{\rm H} \cdot \ma{B}^{\rm H}\right\}$ into ${\rm vec}\left\{\left(\ma{A} \cdot \ma{X}\right)^{\rm H}\right\}^{\rm H} \cdot {\rm vec}\left\{\ma{R} \cdot \ma{X}^{\rm H} \cdot \ma{B}^{\rm H}\right\}$ and apply the rules for generic linear forms twice.