Factorization of Normal Operator

Question

Here is a theorem on Treil's textbook:

Theorem $2.4.$ Any normal operator $N$ in a complex vector space has an orthonormal basis of eigenvectors. In other words, $N=UDU^*$, where $U$ is a unitary matrix, and $D$ is a diagonal matrix.

My question is:

Must $D$ have the eigenvalues of $N$ on its main diagonals?

I assume yes, but do not quite understand.

Answer 1

Yes, and (for this part of the statement) the fact that $N$ is normal is irrelevant. We have $$NU=UD\ ,$$ and extracting the $k$th column on both sides gives $$N{\bf u}_k=\lambda_k{\bf u}_k\ .$$

Answer 2

Yes, the diagonal entries of $D$ are precisely the eigenvalues of $N$. Note that for $\lambda\in\mathbb C$, the operator $$UDU^*-\lambda=U(D-\lambda)U^*$$ is invertible if and only if $D-\lambda$ is invertible, and this happens if and only if $\lambda$ is not on the diagonal of $D$.