Problem from Roman: a lower bound for trace of |T|^2 I'm working on a problem from Stephen Roman's Linear Algebra text, #20 on p. 235: Suppose $\tau \in \mathcal{L}(\mathbb{C}^n)$ and let the characteristic polynomial $\chi_{\tau}(x)$ have roots $\lambda_1, \dotsc, \lambda_n$ (with multiplicity). Show that $$\sum_{i=1}^n|\lambda_i|^2\leq \text{Tr}(\tau^*\tau),$$with equality holding iff $\tau$ is normal.
The last part is clear, since if $\tau$ is normal, then there is an ON basis $B$ diagonalizing $\tau$ and then $[\tau^* \tau]_B$ is diagonal with entries $|\lambda_i|^2$. 
I'm trying to solve the first part. I note that $\text{Tr}(\tau^*\tau)$ is $\sum\limits_{i,j}|a_{ij}|^2$, if $(a_{ij})=[\tau]_B$ for $B$ an ON basis. I am also trying to exploit that $\tau^*\tau\geq0$. I'm stuck, though. Any advice?

Hint: Matrices over $\mathbb{C}$ can be unitarily triangularised. So, you may assume that the matrix of $\tau$ is lower triangular.
