Suppose $A=(a_{ij})$ is positive definite with $a_{ij}\leqslant 0\ (i\neq j)$. Is the matrix $-A+2\mathrm{diag}(a_{11},\dots,a_{nn})$ positive definite?
My attempt: I have shown that the proposition is true for $n\leqslant 3$, using Hurwitz criterion for positive definite matrices. But I have no idea about bigger $n$.
Maybe another way is to show $\lambda_i<2a_{ii}$, where $\lambda_i$ are eigenvalues, but I have no progress.
Any hints/counterexamples are welcomed. THANKS!
The answer for your question is yes and it relies on Perron-Frobenius theorem for non-negative matrices.
First, let us assume that $\text{diagonal}(a_{11},\ldots,a_{nn})=Id$. So $A=Id-C$, where $C_{ii}=0$ and $C_{ij}\geq 0$.
By Perron-Frobenius theorem for non-negative matrices there is an eigenvalue $c\geq 0$ of $C$ such that $c\geq|\lambda|$ for any other eigenvalue $\lambda$ of $C$.
Since $A=Id-C$ is positive definite, we have $c<1$. Therefore $|\lambda|<1$ for any eigenvalue $\lambda$ of $C$.
Therefore $2Id-A=2Id-(Id-C)=Id+C$ is positive definite too.
Now, we can reduce the general case to the previous case by just noticing that if $D=\text{diagonal}(a_{11},\ldots,a_{nn})$ then $2Id-D^{-\frac{1}{2}}AD^{-\frac{1}{2}}$ is positive definite iff $2D-A$ is positive definite.
