Can we provide a bound of the inverse of a matrix plus identify?

Asked Oct 31 '22 at 13:06

Active Oct 31 '22 at 14:40

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I have a similar problem to this question, but instead of finding the exact solution to $(A + I)^{-1}$ where $A$ is a symmetric PSD matrix, I need to somehow find an upper bound for $(A + I)^{-1}$. I am wondering if such a bound is possible to compute?

What I am ideally looking for is an element-wise upper bound on $A^{-1}$. So for example if $\;(A + cI)^{-1} \leqslant A^{-1}$ holds for all $\,c\,$, that would suffice. Moreover, does this generalise beyond the identity matrix, e.g. $\,(A + B)^{-1}\leqslant A^{-1}$ for any PSD matrix $B$ ?

edited Oct 31 '22 at 14:40

asked Oct 31 '22 at 13:06

Qazaz

Hello! Do you mean an upper bound in norm? What have you tried? – G. Gare Oct 31 '22 at 13:18
An upper bound of the norm can be found using the positive eigenvalues of $A+I$, that are the reciprocal of those of $(A+I)^{-1}$ – G. Gare Oct 31 '22 at 13:21
Sorry I should have added this, I have updated the question. – Qazaz Oct 31 '22 at 14:40
It is indeed true that if $(A + B)^{-1} \leq A^{-1}$ – Ben Grossmann Oct 31 '22 at 16:06
Under what assumptions about B? PSD? Or can this be even more general? – Qazaz Oct 31 '22 at 17:36
@Q $A$ needs to be positive definition and $B$ needs to be positive semidefinite. – Ben Grossmann Oct 31 '22 at 19:34
You cannot expect $(A+cI)^{-1}\le A^{-1}.$ For $c$ such that $A+cI>0$ the inequality holds if and only if $c\ge 0.$ So it does not hold for $-|A^{-1}|^{-1}<c<0.$ – Ryszard Szwarc Oct 31 '22 at 20:29
@BenGrossmann Is there a proof for this you could point me to? – Qazaz Nov 02 '22 at 15:01
Can the bound be tighter if $B = cI$? – Qazaz Nov 02 '22 at 15:02
@Qazaz I don't have a reference or link, but in brief: $$ P \geq Q \implies \ I \geq P^{-1/2}QP^{-1/2} \implies\ I \leq P^{1/2}Q^{-1}P^{1/2} \implies\ P^{-1} \leq Q^{-1}. $$ – Ben Grossmann Nov 02 '22 at 17:09
@Qazaz see also this post – Ben Grossmann Nov 02 '22 at 17:22
@Qazaz I just realized I misinterpreted your question: $(A +B)^{-1} \leq A^{-1}$ works as a Loewner bound, not necessarily as an entrywise upper bound. We know that $A^{-1} - (A + B)^{-1}$ is positive semidefinite, but not that it has non-negative entries. – Ben Grossmann Nov 02 '22 at 20:13

Can we provide a bound of the inverse of a matrix plus identify?

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