I just can't show that a unitary matrix $U$ is unitarily diagonizable. I know I need to show that $U$ is unitarily similar to a diagonal matrix, and this result is presumably a consequence of the spectral theorem.
EDIT: I was reading this wrong, and I am supposed to be proving this result without the use of the spectral theorem. I have written a proof using the notion of block matrices which I believe is correct. Can anyone help me prove this result without the spectral theorem, using inner products? I suppose the question reduces to:
Prove that a normal matrix is unitarily diagonalizable (using inner products).
Best Answer
We shall show that unitary matrices are normal, from which the Spectral theorem shall directly apply. The defining property of a normal matrix is $TT^\ast=T^\ast T$. Since $TT^\ast=I$ for $T$ unitary, clearly $T^\ast T=I$ as well, and $TT^\ast=I=T^\ast T$.
If you are using the inner product definition: Recall that $T$ is unitary iff $\langle Tv,Tw\rangle=\langle v,w\rangle$ for every $v,w$ in your inner product space $V$, as in general, $\langle Tv,Tw\rangle=\langle v,T^\ast Tw\rangle=\langle v,w\rangle$ if $TT^\ast=I=T^\ast T$. $T$ normal iff $\langle Tv,Tw\rangle=\langle T^\ast v,T^\ast w\rangle$, and $T$ unitary clearly has this property.