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!