Sum of squared eigenvalues is at most trace of adjoint product Specifically, I'm trying to solve the following:

Let $T$ be a complex $n\times n$ matrix. Let $\lambda_1,\cdots,\lambda_n$ be the eigenvalues of $T$, where each eigenvalue is repeated a number of times equal to its algebraic multiplicity. Prove that
  $$\sum_{k=1}^n|\lambda_k|^2\leq\operatorname{tr}(T^*T),$$
  with equality if and only if $T$ is normal.

It seems like this should be a straightforward proof using a Schur decomposition, but I'm confused by the inequality. It should be the case that $\operatorname{tr}(T^*T)=\operatorname{tr}(TT^*)$, even if $T$ isn't normal, right? So how can this inequality be strict?

Alternatively, we can use the Frobenius inner product: $\langle A,B\rangle=\operatorname{tr}(A^*B)$.
Replace $T$ with $T=U(D+N)U^*$ for some unitary $U$, diagonal $D$, and strictly upper triangular nilpotent $N$ (Schur's theorem). Then
\begin{align*}
    \operatorname{tr}(T^*T)&=\langle T,T\rangle\\
    &=\langle U(D+N)U^*,U(D+N)U^*\rangle\\
    &=\langle D+N,D+N\rangle&\mbox{(because unitary matrices preserve inner products)}\\
    &=\|D\|^2+\|N\|^2+\underbrace{\langle D,N\rangle}_{=0}+\underbrace{\langle N,D\rangle}_{=0}.&\mbox{(by definition of $D$ and $N$)}\\
\end{align*}
Since $\|N\|^2=0$ if and only if $T$ is diagonalizable and $T$ is diagonalizable if and only if $T$ is normal (spectral theorem), the desired result follows from the fact that the entries of $D$ are the eigenvalues of $T$.
