IMC 2014 · Day 2 · P7

Eigenvalues of a real symmetric matrix are real, hence the inequality makes sense. Similarly, for Hermitian matrices diagonal entries as well as eigenvalues have to be real.

Since the trace of a matrix is the sum of its eigenvalues, for $A$ we have $$\sum_{i=1}^{n} a_{ii} = \sum_{i=1}^{n} \lambda_i,$$ and consequently $$\sum_{i=1}^{n} a_{ii}^2 + 2 \sum_{i<j} a_{ii} a_{jj} = \sum_{i=1}^{n} \lambda_i^2 + 2 \sum_{i<j} \lambda_i \lambda_j.$$ Therefore our inequality is equivalent to $$\sum_{i=1}^{n} a_{ii}^2 \le \sum_{i=1}^{n} \lambda_i^2.$$ Matrix $A^2$, which is equal to $A^T A$ (or $A^* A$ in Hermitian case), has eigenvalues $\lambda_1^2, \lambda_2^2, \dots, \lambda_n^2$. On the other hand, the trace of $A^T A$ gives the square of the Frobenius norm of $A$, so we have $$\sum_{i=1}^{n} a_{ii}^2 \le \sum_{i,j=1}^{n} |a_{ij}|^2 = \operatorname{tr}(A^T A) = \operatorname{tr}(A^2) = \sum_{i=1}^{n} \lambda_i^2.$$ The inequality follows, and it is clear that the equality holds for diagonal matrices only.

Remark. Same statement is true for Hermitian matrices.