I've been working on this problem, and I think that I almost have the solution, but I'm not quite there.
Suppose that $A \in M_n(\mathbb C)$ has $n$ distinct eigenvalues $\lambda_1… \lambda_n$. Show that $\sqrt{\sum _{j=1}^{n} \left | {\lambda_j} \right | } \leq \left \| A \right \|_F$.
I tried using the Schur decomposition of $A$ and got that $\left \| A \right \|_F = \sqrt{TT^*}$, but I'm not sure how to relate this back to eigenvalues and where the inequality comes from.
Best Answer
Are you sure that on the left is not the absolute value of the sum? Otherwise, what is the interpretation when all eigenvalues are negative?
In case it is the absolute value of the sum, here's a quick idea when everything's real: the sum of the eigenvalues is the trace of A. So (sum of eigenvalues)$^2$ = (trace of A)$^2\le$ (trace of $A^TA$). Taking square roots of both sides, we find |sum of eigenvalues|$\le\|A\|_F.$