Can any Hermitian positive semi-definite (PSD) matrix $H$ be written as $D + R$, where $D$ is a PSD diagonal matrix and $R$ is a Hermitian PSD matrix with $\operatorname{rank}(R) < \operatorname{rank}(H)$? If not, what are the restrictions on H such that it is possible?
As a follow-up question: if we have an $H$ for which this is possible, when does the lower-rank matrix $R$ also satisfy the condition that it can be decomposed as $R=R'+D'$ with a lower-rank $R'$? Essentially, I would like to create a "chain" of such decompositions such that in the end I have $H=R_1+D$ with some diagonal $D$ and $\operatorname{rank}(R_1)=1$. When is this possible?
Let $e_1,\dots,e_n$ denote the standard basis. A sufficient condition is that if $H x = e_i$ has a solution for some $i$, then there exists a $t > 0$ such that $H - te_ie_i^T$ will be positive semidefinite of a lower rank.
I believe that this condition is also necessary, but I'm not sure.
Proof of necessity: (work in progress) Suppose that $H$ is such that $H - D$ is positive semidefinite for some non-zero diagonal $D$. That is, the matrix $$ H- \sum_{i=1}^n d_i e_ie_i^T $$ is positive semidefinite. So, if $d_k \neq 0$, then $H - d_k e_ke_k^T$ is positive semidefinite. This means that for $0 < t < d_k$, $H - t e_ke_k^T$ is positive semidefinite.
(Not sure about this step:) we may therefore state that $H - t e_ke_k^T$ is PSD for $t \in [0,t_{max}]$ for some $t_{max}$, and that $H - t_{max}e_ke_k^T$ is necessarily of lower rank.