# Inequality involving eigenvalues and trace of an operator with its adjoint

I have come across a problem which I can't properly solve:

Let $\tau$ be a linear operator on $\mathbb{C}^n$ and let $\lambda_{1},...,\lambda_{n}$ be the eigenvalues of $\tau$, each one written a number of times equal to its algebraic multiplicity. I should show that:

$\sum_{i}|\lambda_{i}|^2\leq tr(\tau^*\tau)$

Also, one should show that the equality holds iff $\tau$ is normal.

First I felt that this might use singular values, but I have no success with this. My idea then was that Cauchy-Schwarz may be useful. (I work with matrices, this is clearly not a restriction to the problem.) So I defined the inner-product $\langle A,B\rangle=tr(B^*A)$, which I know to be acceptable. Elementary operations on Cauchy-Schwarz inequality

$|\langle A,A^*\rangle|^2$$\leq \langle A,A\rangle\langle A^*,A^*\rangle then give that |\sum_{i}\lambda_{i}^2|^2$$\leq (tr(A^*A))^2$ (I may be mistaken). This is not what I want. In the question ''equality holds iff $\tau$ is normal'', one way (right to left) is easy.

I highly appreciate any suggestion!

-

Let's do a simpler thing.

We have a Schur decomposition $\tau=u^*\sigma u$ with $u$ unitary and $\sigma$ an upper triangular matrix. Since both sides of the inequality take the same value for $\tau$ and for $\sigma$ we may assume that $\tau$ is in fact itself upper triangular. Now the inequality is obvious, because the right hand side is a sum of the left hand side and some other, non-negative terms.

-
I fail to follow your argument. For diagonal matrices, the inequality in the question is an equality. Why and how would that extend as a (proper) inequality to more general matrices? –  Martin Argerami Sep 6 '12 at 1:46
As an example of where the inequality is strict, consider the matrix representation of the right shift operator $R$, $e_1 \rightarrow e_2 \rightarrow e_3 \rightarrow 0$ and it's adjoint the left shift operator $L=R^*$, $e_3 \rightarrow e_2 \rightarrow e_1 \rightarrow 0$. The composition $RL$ acts as the identity on $e_1$ and $e_2$, and sends $e_3$ to zero, so it has trace 2, whereas there are no eigenvalues of $R$ or $L$. –  Nick Alger Sep 6 '12 at 6:03
N.B. I have just edited my answer, and the two comments aobve refer to the old version. –  Mariano Suárez-Alvarez Sep 6 '12 at 6:27

It wouldn't surprise me if I'm missing something, but I couldn't think of a straightforward proof. But this inequality is a particular case of Weyl's Majorant Theorem (see Theorem II.3.6 in Bhatia's Matrix Analysis, or there are surely many other references).

-
I agree with you about my retracted "answer." –  Kevin Carlson Sep 6 '12 at 5:10
Thank you a lot. –  Lucien Sep 6 '12 at 9:10
You are welcome. I couldn't address the implication "left implies right" with this idea. But it does follow easily from Mariano's argument (the case of equality forces $\sigma$ to be diagonal). –  Martin Argerami Sep 6 '12 at 14:17