I see that this is not true in general.
But it is true in hermitian matrix. How to prove that this is true?
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1The Schur-Horn theorem is related, but doesn't quite help us here. – Ben Grossmann Oct 05 '15 at 06:32
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1This is a direct consequence of Thompson-Sing theorem (see thm 2.4 of Chi-Kwong Li's notes), but the theorem is perhaps an overkill. – user1551 Oct 05 '15 at 13:51
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The result is false for some complex matrices, eg. $\begin{pmatrix}5&-4\\6&-5\end{pmatrix}$; yet, it is true for any normal matrix.
Proof. Let $A=[a_{i,j}]$ be a normal matrix and $spectrum(A)=(\lambda_i)_i$. There is $U$ (unitary) s.t. $A=Udiag((\lambda_i)_i)U^*$. Then, for every $i$, $a_{i,i}=\sum_{j} \lambda_j u_{i,j}\overline{u_{i,j}}$. Thus $\sum_i |a_{i,i}|\leq \sum_j|\lambda_j|\sum_i |u_{i,j}|^2=\sum_j|\lambda_j|$.
We can do better (cf. user1551 post). For any matrix $A$, $\sum_i |a_{i,i}|\leq \sum_i s_i$ where the $(s_i)_i$ are the singular values of $A$.
Proof. There are unitary $U,V$ s.t. $A=Udiag((s_i)_i)V$; let $a_{j,j}=r_jexp(i\theta_j)$ and $D=diag(exp(-i\theta_j)_j)$. One has $DA=DUdiag(s_i)V$ or $A'=Wdiag(s_i)V$ where the elements of the diagonal of $A'$ are the $|a_{i,i}|$ and $W$ is unitary. Thus $|a_{i,i}|=\sum_j w_{i,j}s_jv_{j,i}$ and $\sum_i|a_{i,i}|=\sum_js_j\sum_iw_{i,j}v_{j,i}\leq \sum_js_j$ by Cauchy-Schwartz.
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Nice and simple! +1. And I like that you tried to prove things in weaker settings. – user1551 Oct 06 '15 at 08:34
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