Is the commutator of two unitary operators normal?

Question

Given two dxd unitary operators $U,V$, is $[U,V]$ normal? I was able to show it for the d=2 case by direct calculation, but not sure about the general case.

Answer 1

Thoughts: Hopefully this leads someone else to an answer

We find that $$ [U,V]^* = (UV - VU)^* = V^*U^* - U^*V^* $$ So that $$ [U,V][U,V]^* = (UV - VU)(V^*U^* - U^*V^*) =\\ 2I - UVU^*V^* - VUV^*U^* $$ Similarly, calculate $$ [U,V]^*[U,V] = 2I - U^*V^*UV - V^*U^*VU $$ So, $[U,V]$ will be normal iff $$ UVU^*V^* + VUV^*U^* = U^*V^*UV + V^*U^*VU \iff\\ [UVU^*V^*] + [UVU^*V^*]^* = [U^*V^*UV] + [U^*V^*UV]^* \iff\\ [UVU^*V^* - U^*V^*UV] = [V^*U^*VU - VUV^*U^*] \iff\\ [UV,U^*V^*] = [V^*U^*,VU] \iff\\ [UV,U^*V^*] = - [UV,U^*V^*]^* $$

Answer 2

I tried this on random 3x3 unitary matrices and found many pairs whose commutators aren't normal. In particular, this is true for the following pair:

$$ \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & 1& 1\\ 1& e^{i 2\pi/3} & e^{-i2 \pi/3}\\ 1 & e^{-i2 \pi/3} & e^{i 2 \pi/3} \end{bmatrix}, \frac{1}{\sqrt{3}} \begin{bmatrix} 1 +i & \frac{1}{\sqrt{2}}(1-i)& 0\\ -\frac{i}{\sqrt{2}}& 1 & \frac{1}{\sqrt{2}}+i\\ \frac{1}{\sqrt{2}} & i & 1-\frac{i}{\sqrt{2}} \end{bmatrix}, $$