Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A^{-1}$.

share|cite|improve this question
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
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 $\|A^{-1}\| \le 2/\delta$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.