# Absoute sum of diagonal elements no more than absolute sum of eigenvalues

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.

This isn't true. Every nilpotent matrix $$A$$ with a nonzero diagonal element can serve as a counterexample, such as $$A=\pmatrix{1&-1\\ 1&-1}$$.
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.