Let $A\geq 0$ (i.e., $x^*Ax\geq 0$, $\forall x\in\mathbb{C}^n$) be a PSD matrix. Given $A$, is it always possible to find a digonal $D\geq 0$ and $D\neq0$ such that $$A-D\geq 0?$$
I think it could be possible unless $A$ is a rank-one Idempotent matrix. Note that if $A$ is full rank, then $D=\lambda_{\min}(A)\mathbb{I}$ is a possible choice. But don't know when $1<$rank($A$)$<n$.
No, this won't work if $A$ is singular, even if the rank has the maximal value $n-1$. Take any normalized vector $v$ with no zero entries, then choose an ONB $\{v, w_2, \ldots ,w_n\}$ and take $A=\sum_{j=2}^n w_j w_j^*$. Then $$ \langle v, (A-D)v\rangle = -\langle v, Dv \rangle = -\sum d_j |v_j|^2 < 0 . $$