Showing that $UU^*=D$ where $D$ is a diagonal matrix

Question

I've been receiving downvotes and would like to know why?

pretty much the title. I think this condition isn't generally true but maybe I'm missing something (my maths could be horribly wrong). I was just wondering if someone wiser had any insights or could steer me in the right direction. (This image is from QFT for the gifted amateur incidentally). I have been reading around quite a bit and some authors choose not to use this proof and instead demonstrate that there are instances where the time reversal operator squared will be 1 and others where it'll be -1. But this proof seems elegant to me so I just wanted to make sure I wasn't missing something obvious.

Answer 1

Recall that if $U$ is unitary, then $U^\dagger = U^{-1}$. Using this, $D = UU^*$ satisfies

$$D^\dagger D = (UU^*)^\dagger (UU^*) = U^TU^\dagger UU^* = U^T U^* = (U^\dagger U)^* = \mathbf{1}.$$

Hence, $D$ is unitary and diagonalizable by the spectral theorem.