Normal Matrices Unitarily Diagonazible

Question

Disclaimer: This thread is just meant to record (Q&A).

Are the unitarily diagonazible matrices precisely the normal ones?

Surely, every normal matrix has an eigenbasis.
Now I got asked by a friend wether the converse holds true as well.

Answer 1

Yes, it is true! This can be seen most easily starting from its diagonalization exploiting the unitarity: $$\quad A^*A=(U^{-1}D^*U)(U^{-1}DU)=U^{-1}D^*DU=U^{-1}DD^*U=(U^{-1}DU)(U^{-1}D^*U)=AA^*$$