Denote $A_{\text{off-diag}}:=A-\text{diag}(A)$, i.e. setting diagonal elements to be zero. Denote $A\succneqq0$ iff $A$ is positive semidefinite (and not positive definite), while $A\succ0$ iff $A$ is positive definite.
Can we prove that $\not\exists A\succneqq0$ such that $A+A_{\text{off-diag}}\succ0$?
Here is what I thought. $A+A_{\text{off-diag}}$ scales up the off-diagonal elements, making $\text{diag}(A)$ less dominant. If $A$ is not positive definite already, then scaling up the diagonal elements will make it even "less positive definite".
I appreciate a proof (or a hint) or a counterexample. Thanks!
Hint: Suppose there exists an $A \succneqq 0$ such that $A+A_{\text{off diag}} \succ 0$. Since $A \succneqq0$, then there exists an eigenvector $v \neq 0$ corresponding to the eigenvalue $0$. This satisfies $v^TAv = 0$. Since $A+A_{\text{off diag}} \succ 0$, we must have $v^T(A+A_{\text{off diag}})v > 0$. Now what can you say about $2v^TAv - v^T(A+A_{\text{off diag}})v$?