Unitary equivalence and eigenvalues After a search I did find threads with similar topics, but none with exactly what I want to know.
A matrix $A$ is complex and normal [real and symmetric] if and only if it is unitarily [orthogonally] equivalent to a [real] diagonal matrix.
Write $A$ as $A=P^*DP$.  Then are the entries of $D$ the eigenvalues of $A$?
Let's say $A-tI = Q^*D'Q$.  I believe this is still complex and normal [real and symmetric].  Of course we have $\det(A-tI)=\det(Q^*DQ)=\det(Q^*)\det(D')\det(Q)=\det(D')$, and this does the trick if the entries of $D'$ are $A_{ii}-t$, but I guess it's not clear to me that this is the case.
I suspect this is true, because if it is then some nice things that I'm trying to prove follow from it (wishful thinking, I know).  So how do we prove it?

The diagonal entries of $D$ are the eigenvalues of $A$, yes. 
Simply note the eigenvectors of $D$ are the standard vectors, with eigenvalues the diagonal entries. And if $\mathbf{v}_i = P^*\mathbf{e}_i$, then
$$A\mathbf{v}_i = AP^*\mathbf{e}_i = (P^*DP)P^*\mathbf{e}_i = P^*D\mathbf{e}_i = P^*d_{ii}\mathbf{e}_i = d_{ii}(P^*\mathbf{e}_i),$$
so $P^*\mathbf{e}_i$ is an eigenvector of $A$ with eigenvalue $d_{ii}$ (the $i$th entry in the diagonal of $D$). Sincee $P^*$ invertible, the vectors $P^*\mathbf{e}_i$ are linearly independent, so this shows $A$ and $D$ have the same eigenvalues with the same multiplicities. And of course, the eigenvalues of $D$ are just the diagonal entries of $D$.
(Note: the argument above applies to any relation by conjugation: if $A=M^{-1}BM$, then $A$ and $B$ have the same eigenvalues with the same multiplicities, since if $\mathbf{v}$ is an eigenvalue of $B$, then $M^{-1}\mathbf{v}$ is an eigenvalue of $A$.)
