A well known result by Varah states that if $A$ is a strictly diagonally dominant matrix of dimension $n$, then $\|A^{-1}\|_{\infty} \le \max_i\frac{1}{|a_{kk}|-\sum_{j \neq k}|a_{kj}|}$, where the infinity norm here is the induced $\ell_{\infty}$, i.e. $\max_{k}\sum_{j}|a_{kj}|$.

Let us assume that $A$ is real and Hermitian, with eigenvalues in $[-a,a]$, all distinct. I am trying to bound the infinity norm of the following resolvent:

$$ B = \left( \delta i I - A \right)^{-1} $$

where $\delta$ is small, so $\delta i I -A$ is not necessarily diagonally dominant. Naturally, $\delta i I -A$ is always invertible. I denote $i$ as the imaginary unit and $I$ the identity matrix.

A trivial upper bound would be $\frac{\sqrt{n}}{\delta}$. Do you see any circumstances for which a better bound may be attained?

Thank you.

Edit: A bound of $\epsilon$ to the above question is equivalent to asking whether for all matrices $E$ with $\|E\|_{\infty} \le \epsilon$, $i\delta$ is not an eigenvalue of $A+E$.

Generally, $\epsilon$ will be a function of both $\delta$, $\|E\|$ and the condition number of the eigenvalue matrix of $A$. However, subject to the above restrictions, do you see a better bound?

  • $\begingroup$ What is $i$? The imaginary unit? The identity matrix? $\endgroup$ – Federico Poloni Apr 21 '14 at 18:00
  • $\begingroup$ @FedericoPoloni I'm pretty sure from context it is the imaginary unit (multiplied by a copy of the identity matrix of the appropriate size) $\endgroup$ – Yemon Choi Apr 21 '14 at 18:14
  • $\begingroup$ @neil What other structure are you willing to put on your matrix $A$ (beyond trivial things like "being diagonal with real entries") in order to get "a better bound"? Generically I would think that since $\Vert A \Vert_\infty$ can be significantly bigger than $\rho(A)$ even when $A$ is Hermitian, the same phenomenon would occur for $\Vert B \Vert_\infty$ $\endgroup$ – Yemon Choi Apr 21 '14 at 18:20
  • $\begingroup$ @FedericoPoloni, i is the imaginary unit. Thank you, I edited. $\endgroup$ – neil Apr 21 '14 at 19:20
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    $\begingroup$ @YemonChoi, I am also willing to assume $A$ is a "shifted" doubly stochastic matrix. Namely, that $A = S -\alpha I$ where $A$ is stochastic and $\alpha$ is some real number in $[0,1]$. $\endgroup$ – neil Apr 21 '14 at 19:22

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