… 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 […]