Subtracting a diagonal PSD matrix from a given PSD matrix

Question

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($A$)$<n$.

Thanks for your valuable(!) and very insightful(?) comment!! — Poor.Jack, Jun 20 '16 at 15:39
Now, if you assume $D\neq 0$, then what $D$ will you take for $A=0$? — Sungjin Kim, Jun 20 '16 at 15:43

score 1 · Accepted Answer · answered Jun 20 '16 at 18:22

1

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 $$ \langle v, (A-D)v\rangle = -\langle v, Dv \rangle = -\sum d_j |v_j|^2 < 0 . $$

answered Jun 20 '16 at 18:22

Great! Thanks a lot, for your help. I was about to delete this question, but then was waiting for particularly your thought. – Poor.Jack Jun 20 '16 at 18:25
Could you please share your thought on my new problem. – Poor.Jack Jul 29 '16 at 02:21

Subtracting a diagonal PSD matrix from a given PSD matrix

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