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?