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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?

Atsina
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    As to equality: If $T$ is normal, then there exists a orthonormal basis $(e_i)$ of eigenvectors of $T$, hence $tr(T^T) = \sum_i\langle T^Te_i,e_i\rangle = \sum_i|Te_i|^2 = \sum_i|\lambda_i|^2$. – amsmath Oct 09 '19 at 02:21

2 Answers2

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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.

Martin Argerami
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  • And equality obviously holds if and only if $Te_j = \lambda_je_j$ for all $j$, which means that $T$ is normal (and hence also the original $T$ is normal). – amsmath Oct 09 '19 at 02:54
  • Indeed. I'll mention that. – Martin Argerami Oct 09 '19 at 03:00
  • I hastily accepted your answer...could you elaborate on how you got from the 2nd line to the 3rd line? Also why is it clear that the triangularization of $T$ gives the same eigenvalues as $T$ with respect to the standard basis? – Atsina Oct 28 '19 at 18:37
  • I assume you mean $|Te_j|^2=\langle Te_j,Te_j\rangle=\langle T^*Te_j,e_j\rangle$? And $|e_j|=1$ because ${e_1,\ldots,e_n}$ is orthonormal. – Martin Argerami Oct 28 '19 at 19:29
  • As for the triangularization, it looks like you had missed that I mentioned a unitary conjugation. But I changed that part a bit now. – Martin Argerami Oct 28 '19 at 19:31
  • Okay, but $(e_j)$ are not eigenvectors unless $T$ is diagonal. So the first and last equalities seem like an issue. – Atsina Oct 30 '19 at 17:05
  • I never said they were eigenvectors. Have you tried calculating $\langle Te_j,e_j\rangle$ for a matrix $T$ that is triangular with respect to the basis ${e_j}$? And for the last equality, do you know what the trace is? – Martin Argerami Oct 30 '19 at 17:09
  • I thought you were taking sum of eigenvalues, but I see you're actually taking sum on the diagonal. Even still, how can you conclude $T$ is normal without $(e_j)$ being eigenvectors? You seem to be using the spectral theorem – Atsina Oct 30 '19 at 17:18
  • Are you talking about the inequality or the equality? The inequalities hold for any $T$, that's why one needs to triangularize instead of diagonalize. The last paragraph refers to the case of equality; equality forces $T$ to be diagonal in the basis ${e_j}$, so $T$ is normal. So yes, equality forces the $e_j$ to be eigenvectors. – Martin Argerami Oct 30 '19 at 17:21
  • So just to be clear, the 1st equality comes from the fact that the sum of the diagonal of a triangular matrix equals the sum of the eigenvalues, the 2nd (in)equality is Cauchy Schwarz, the 3rd (in)equality is definition, and the last (in)equality is again sum of diagonal, right? And then equality iff $T$ is diagonalizable iff $T$ is normal – Atsina Oct 30 '19 at 17:30
  • Almost. In a triangular matrix, the diagonal entries are the eigenvalues. And yes, the last equality is the definition of the trace, which doesn't depend on the orthonormal basis one uses. – Martin Argerami Oct 30 '19 at 17:50
  • Okay I'm almost ready to accept your answer again...could you just help me understand why the first and last inner products you use give the diagonal entries? I can see this by computing examples, but what's the theory behind it? – Atsina Oct 30 '19 at 18:18
  • By definition, the $k,j$ entry of $T$ in the basis ${e_j}$ is $T_{kj}=\langle Te_j,e_k\rangle$. – Martin Argerami Oct 30 '19 at 18:27
  • Please consider upvoting, too. Not just my answer, but all you find helpful. That's how the site works. – Martin Argerami Oct 30 '19 at 18:44
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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 $T$.

Atsina
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