Sum of squared eigenvalues is at most trace of adjoint product

Question

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?

Answer 1

For an easy example where the inequality is strict, take $T=\begin{bmatrix} 0&1\\0&0\end{bmatrix}$. Then $$ \sum_{j=1}^2|\lambda_j|^2=0+0<1=\operatorname{Tr}\left(\begin{bmatrix} 0&0\\0&1\end{bmatrix}\right)=\operatorname{Tr}(T^*T). $$ As for the inequality, use the Schur decomposition to get an orthonormal basis $\{e_1,\ldots,e_n\}$ such that $T$ is triangular. Then \begin{align} \sum_{j=1}^n|\lambda_j|^2 &=\sum_{j=1}^n|\langle Te_j,e_j\rangle|^2\\ &\leq\sum_{j=1}^n\|Te_j\|^2\,\|e_j\|^2\\ &=\sum_{j=1}^n\langle T^*Te_j,e_j\rangle\\ &=\operatorname{Tr}(T^*T). \end{align}

As mentioned by amsmath, equality occurs precisely when we have equality in Cauchy-Schwars in the second line. This means that $Te_j$ and $e_j$ have to be colinear for all $j$, i.e. there exist numbers $\lambda_j$ with $Te_j=\lambda_je_j$. Then $T$ is normal.

Answer 2

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 $T$.