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Let $A$ be a Hermitian matrix and $\lambda_\max(A)$ be its maximal eigenvalue. Suppose $B$ is the matrix produced from $A$ by increasing the magnitude of any element of $A$ along with its diagonally opposite element so that $B$ remains Hermitian. That is, if we consider $A(i,j)$ and change it to say $a+\mathtt{i}b$ such that $|a+\mathtt{i}b|>|A(i,j)|$, then $$B(k,l)=\begin{cases}A(k,l)& \text{ if } (k,l)\ne(i,j)\text{ or }(k,l)\ne(j,i),\\a+\mathtt{i}b & \text{ if } (k,l)=(i,j),\\a-\mathtt{i}b&\text{ if } (k,l)=(j,i).\end{cases}$$ My observations makes the hypothesis that $$\lambda_\max(A)\le\lambda_\max(B).$$ Is it true? How to prove it?

Note that the above hypothesis can be thought of as a generalization of the statement:

The increase of any element of a nonnegative matrix does not decrease its maximal eigenvalue.

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This isn't true. User tmyklebu has exhibited a counterexample in an answer to another question: for $$ A=\pmatrix{2&1&1\\ 1&2&0\\ 1&0&2}, \ B=\pmatrix{2&1&1\\ 1&2&-1\\ 1&-1&2}, $$ we have $\rho(A)=2+\sqrt{2}>\rho(B)=3$.

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  • $\begingroup$ And of course if you go from $A$ to $B$ in two steps, each changing only one entry, then one of the steps will also be a counterexample.) $\endgroup$
    – GEdgar
    Aug 30 '17 at 13:26

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