# Frobenius Norm and Relation to Eigenvalues

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 $$\sqrt{\sum _{j=1}^{n} \left | {\lambda_j} \right |^2 } \leq \left \| A \right \|_F\,.$$

I tried using the Schur decomposition of $$A$$ and got that $$\left \| A \right \|_F = \sqrt{TT^*}$$, where $$A=QTQ^*$$ with $$Q$$ unitary and $$T$$ triangular, but I'm not sure how to relate this back to eigenvalues and where the inequality comes from.

You are in the right way. The corresponding Schur decomposition is $A = Q U Q^*$, where $Q$ is unitary and $U$ is an upper triangular matrix, whose diagonal corresponds to the set of eigenvalues of $A$ (because $A$ and $U$ are similar). Now because Frobenius norm is invariant under unitary matrix multiplication:

$$||QA||_F = \sqrt{\text{tr}((QA)^*(QA))} = \sqrt{\text{tr}(A^*Q^* QA)} = \sqrt{\text{tr}(A^*A)} = ||A||_F$$

(the same remains for multiplication of $Q$ on the right) then we could write: $$||A||_F = ||Q U Q^*||_F = ||U||_F \rightarrow \sqrt{\sum_{j=1}^n |\lambda_j|^2} \leq ||A||_F$$

Note: The inequality comes from the definition of the Frobenius norm: The sum of the square of all entries in the matrix. Since $U$ contains the eigenvalues on his diagonal, the term in the left has to be less or equal to the sum over all entries, because $U$ could have non zero entries over his diagonal.
• Took me a while to get what's going on here, but I think it is with equality if $A$ is symmetric, but in general if we instead use the Jordan form (rather than the Schur), then it can have nontrivial offdiagonal values that contribute to the frobenius norm. Commented Jun 4, 2019 at 19:55