TITLE: Sum of the absolute eigenvalues of A>=B
QUESTION [7 upvotes]: Kindly help me to prove/disprove the following statement. 
Let $A$ be a symmetric matrix of order $n \times n$ with all the diagonal entry equal to $0$, and other non-diagonal entry equal to $k$ (where $k$ is a fixed positive integer). Similarly,Let $B$ be a symmetric matrix of order $n \times n$ with every diagonal entry equal to $0$, and each non-diagonal entry equal to some $\ell$ with $1\leq \ell \leq k<n$.
For example:$ A= \begin{pmatrix} \begin{array}{cccc}
 0 & 3 & 3 & 3 \\
 3 & 0 & 3 & 3 \\
 3 & 3 & 0 & 3 \\
 3 & 3 & 3 & 0 \\
\end{array}\end{pmatrix}$, 
$B=\begin{pmatrix}
\begin{array}{cccc}
 0 & 3 & 1 & 2 \\
 3 & 0 & 2 & 3 \\
 1 & 2 & 0 & 1 \\
 2 & 3 & 1 & 0 \\
\end{array}
\end{pmatrix}$.
Prove/disprove the following statement: 
Suppose $\lambda_1, \lambda_2, ..., \lambda_n$ are eigenvalues of $A$ and $\mu_1, \mu_2, ..., \mu_n$ are eigenvalues of $B$. Then  $$\sum_{i=1}^{n}{|\lambda_i|}\geq \sum_{i=1}^{n}{|\mu_i|}.$$ Note that $A$ and $B$ are the distance matrices of some vertices(diameteral) of $G_1$ and $G_2$ respectively.

REPLY [10 votes]: The claim is false.
In particular, we have $A=kee^T-kI$, so that $\lambda(A)=((n-1)k,-k,\ldots,-k)$, so that $\|A\|_* = 2(n-1)k$.
Now generate a random matrix $B$ such that $B_{ii}=0$, $B_{ij}=B_{ji}$ and $B_{ij} \le A_{ij}$ for every entry. It does not matter that the entries of $B$ are integers or not (to have a "clean" example, I round the entries below to ensure that $B$ is integral). In particular, try the following Matlab code:
n=63;k=3;B=ceil(k*rand(n));B=B-diag(diag(B));B=round((B+B')/2);sum(svd(B))-2*(n-1)*k

Quite easily one obtains $\|B\|_* > \|A\|_* = 2(n-1)k=372$.  
Other choices of $n$ and $k$ can give you smaller explicit examples if you wish to find (e.g., $n=23, k=2$ also worked after some attempts).