Sum of squares of eigenvalues This came up in a sample final of mine, but I couldn't figure out a way to solve it.
For any $n \times n$ complex-valued matrix $A$, define $N(A) = \sum_{i,j=1}^n a_{ij}\bar{a_{ij}}.$ Let $A$ be a matrix with the eigenvalues $\{ \lambda_1, ..., \lambda_k \}$. Show that $N(A) \geq \sum_{i=1}^k \lambda_{i}^2.$
I managed to show that equality holds when $A$ is self adjoint, but I was not able to figure out anything further from this. I also tried an approach where $A = UTU^*$, where $U$ is unitary and $T$ is upper-triangular, but again I hit a dead-end. Could anyone help with this? Thank you!
 A: Your idea to try $A = UTU^*$ is good.
Since in $N(A) \geq \sum_{i=1}^k \lambda_{i}^2$ you treat the eigenvalues as members of an ordered field, I think you assume that they are real. We will prove a slightly more general $N(A) \geq \sum_{i=1}^k |\lambda_{i}|^2$ where eigenvalues are allowed to be complex (which may have been your intention all along).
First note that $N(A) = {\rm tr}(AA^*)$. Using your substitution and the cyclic property of the trace $N(A) = {\rm tr}(UTU^*UT^*U^*) = {\rm tr}(TT^*)$. Therefore
$$
N(A) = \sum_{i \leq j} |t_{ij}|^2 = \sum_{i=0}^k |\lambda_i|^2 + \sum_{i \lt j} |t_{ij}|^2 \geq \sum_{i=0}^k |\lambda_i|^2
$$
where we used the fact that diagonal entries of a triangular matrix are its eigenvalues (which in turn follows from properties of the determinant).
A: Let $M_n(\mathbb{C})$ denote the set of all $n\times n$ complex matrices. Also, define
$$ \langle A, B \rangle = \sum_{i,j} a_{ij}\overline{b_{ij}} = \operatorname{Tr}(A B^*) $$
for $A, B \in M_n(\mathbb{C})$. Then it is clear that $\langle \cdot, \cdot \rangle$ defines a Hermitian inner product on $M_n(\mathbb{C})$. So, if $\lambda_1,\dots,\lambda_n$ are eigenvalues of $A \in M_n(\mathbb{C})$, counted with multiplicity, then by the Cauchy-Schwarz inequality,
$$ \Biggl| \sum_{j=1}^{n} \lambda_j^2 \Biggr| = \left| \operatorname{Tr}(A^2) \right| =  \left| \langle A, A^* \rangle \right| \leq \left|\langle A,A \rangle\right|^{1/2}\left|\langle A^*,A^* \rangle\right|^{1/2} = \operatorname{Tr}(AA^*). $$
