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Suppose $A$ is an $n\times n$ matrix such that $A+A^H-\delta I_n$ is positive-semidefinite, for some $\delta>0$, then can we show a bound on the norm of $A^{-1}$ ? Can we show that this the norm of the inverse of $A$ is at most $2/\delta$ ? (The norm is the usual matrix 2-norm)

Let me briefly describe my approach. Please correct me if I am wrong.

By the positive semi-definiteness condition, we obtain that $v^H (A+A^H) v \geq \delta$, $\forall v$ of norm $1$. So the norm of $A+A^H$ is at least $\delta$, and using Cauchy-Schwarz inequality, we obtain a lower bound on the norm of $A$ as $\delta /2$. But this does not seem to help in upper bounding the norm of $A^{-1}$.

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I changed $n*n$ to $n\times n$, coded as n\times n. $\TeX$ is sophisticated. The asterisk for multiplication is a workaround for use when you can only use symbols on the keyboard and "x" must be available as a variable. So it amounts to eating mashed potatoes with your fingers when silverware is right there. Also, I changed $\delta$$>$$0$ to $\delta>0$, with all three symbols inside just one instance of $\TeX$ That way you get proper spacing between characters. –  Michael Hardy Sep 3 '12 at 3:03
Thanks for the edit. I am relatively new to TEX and hope to improve with more usage. –  BharatRam Sep 3 '12 at 3:28
It's sufficient to consider this question for $\delta = 1$. –  Hans Engler Sep 3 '12 at 3:42
@all: I deleted some comments about topology which is irrelevant to the question. –  Willie Wong Sep 3 '12 at 11:36
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1 Answer

up vote 2 down vote accepted

For any nonzero vector $u$, $$\|u\| \|A u\| \ge |u^H A u| \ge \text{Re}(u^H A u) = u^H (A + A^H) u/2 \ge \delta \|u\|^2/2$$ so $\|Au\| \ge \delta \|u\|/2$, which implies that $A$ is invertible with $\|A^{-1}\| \le 2/\delta$.

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This only gives a lower bound on the norm of $A$, as I mentioned as part of my attempt. How did you conclude that $A$ is invertible and upper bound the norm of its inverse? –  BharatRam Sep 3 '12 at 5:51
No, it gives a lower bound on the norm of $A u$ for every nonzero $u$. 1) If a square matrix is invertible, its null space is nonzero, so there would be $u \ne 0$ with $\|A u\| = 0 < \delta \|u\|/2$. 2) Take $u = A^{-1} v$ to get $\|v\| \ge \delta \|A^{-1} v\|/2$. –  Robert Israel Sep 3 '12 at 5:55
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