Unitary equivalence

Question

How to show that: Let $A=QR$ be a $QR$ factorization, then $A$ is normal if and only if $Q$ and $R^*$ can be simultaneously diagonalized by unitary equivalence.

Answer 1

The "if" part is trivial. The "only if" part, however, is not true in general. Here is a counterexample: $$ A=\pmatrix{5&2&-4\\ 2&8&2\\ -4&2&5} =QR= \pmatrix{ \tfrac{ 5}{\sqrt{45}} &0 &\tfrac23\\ \tfrac{ 2}{\sqrt{45}} &\tfrac2{\sqrt{5}} &\tfrac{-1}3\\ \tfrac{-4}{\sqrt{45}} &\tfrac1{\sqrt{5}} &\tfrac23} \pmatrix{ \tfrac5{\sqrt{45}}&\tfrac2{\sqrt{45}}&\tfrac{-4}{\sqrt{45}}\\ 0&\tfrac2{\sqrt{5}}&\tfrac1{\sqrt{5}}\\ 0&0&0} $$ If $Q$ and $R^\ast$ are simultaneously diagonalisable, they must commute, i.e. $R^\ast Q-QR^\ast=0$. Yet, in this example, we have $$ R^\ast Q-QR^\ast=9\sqrt{5}\pmatrix{8&-6&10\\ -4&3&-2\\ 8&-6&-11}\ne0. $$