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?

-
If $A = P^{\ast} DP$, then $A - tI = P^{\ast} (D - tI) P$ and $\det(A - tI) = \det(D - tI)$. More generally, eigenvalues are preserved by conjugation. – Qiaochu Yuan Jan 29 '12 at 5:45

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 $A$.)
I take it the collection of $\mathbb{v}_i$ is the basis in which $A$ is unitarily equivalent to $D$? The only thing I'm not clear on is how we know that $\mathbb{v}_i=P^*\mathbb{e}_i$. – Alex Petzke Jan 29 '12 at 21:38
@Alex: I define $v_i$ to be $P^*e_i$; then I show that the $v_i$ are eigenvectors of $A$. Yes, they are a basis that realizes the unitary equivalence to $D$, because such a basis is given by the columns of $P^*$. But that fact is irrelevant to the computations. – Arturo Magidin Jan 29 '12 at 21:41