Let $A, B$ be two equal-sized, symmetric matrices with $A,B \succeq 0$, and let $\epsilon > 0$. I am conjecturing that: $$A - \epsilon B \succeq 0 \quad\text{as}\quad \epsilon \to 0$$ if (i) $rank (A) \ge rank(B)$, and (ii) all diagonal entries of $A>0$.
My question is whether my conjecture is true and, if yes, how to prove it.
Any help is much appreciated.
EDIT: The question is whether the above holds for small but positive $\epsilon$; i.e., whether there exists some $\bar\epsilon$ such that the condition holds for all $0 < \epsilon < \bar \epsilon$
Your conjecture, if I understand it correctly, is false. Take $$ A = \frac 12 \pmatrix{1&1\\1&1}, \quad B = \pmatrix{1 &0\\0&0} $$ verify that for all $\epsilon > 0$: $\det(A - \epsilon B) < 0$.