Suppose $A$ is an $n\times n$ matrix, and $\lambda_1, \lambda_2, \ldots, \lambda_n$ are its eigenvalues. Prove that
$$ \sum_{i = 1}^n \lvert{A_{ii}} \rvert \leq \sum_{i = 1}^n \lvert \lambda_i\rvert. $$
This should be true because I have verified it on 100,000 $10\times 10$ matrices. However, I failed to prove it since I cannot find a convenient representation of sum of these absolute values.
UPDATE: It is not true for general matrices, but does hold for normal matrices.
On
This does hold if we assume $A$ is normal
Let $D$ be a unitary diagonal matrix where $d_{i,i}$ has the same polar angle as $\bar{a_{i,i}}$
(and if $a_{i,i}=0$ then set $d_{i,i}=1$)
$\sum_{i = 1}^n \lvert{a_{ii}} \rvert = \text{trace}\Big(DA\Big)\leq \sum_{i=1}^n 1 \cdot \sigma_i = \sum_{i = 1}^n \lvert \lambda_i\rvert$
by the von-Neumann trace inequality and the fact that $\sigma_i=\vert \lambda_i\vert$ for normal matrices.
This isn't true. Every nilpotent matrix $A$ with a nonzero diagonal element can serve as a counterexample, such as $A=\pmatrix{1&-1\\ 1&-1}$.