Let $D=diag (d_{ii}) \in M_n(\mathbb R)$ be a diagonal matrix and $E\in M_n(\mathbb R)$ be such that

$||E||_\infty < \min _{i\ne j} \Bigg|\dfrac{d_{ii}-d_{jj}}{2}\Bigg|$.

Then how to show that there is an ordering of the eigenvalues of $D+E$ as $\{\mu_1,...,\mu_n\}$ such that $|d_{ii}-\mu_i|<||E||_\infty $ ?

I think I have to apply Gershgorin circle theorem, but I'm not quite sure how.

Please help

NOTE: Here $||E||_\infty :=\sup_{||x||_\infty=1}||Ex||_\infty$


Hint: For convenience, I'll take $\delta := \min _{i\ne j} |{d_{ii}-d_{jj}}|$.

We note that $$ \|E\|_{\infty} = \max_{i=1,\dots,n} \sum_{j=1}^n |e_{ij}| $$ Thus, when we draw the Gershgorin disks for $D+E$, the center of the $i$th disk will be $d_{ii} + e_{ii}$, and the radius will be $R_i = \sum_{j \neq i} |e_{ij}| < \frac 12 \delta - |e_{ii}|$ (note that we must have $|e_{ii}| \leq \|E\|_\infty < \frac 12 \delta$). I claim that this is enough for us to conclude that the disks will not overlap.

Note that each disk fits inside a disk centered at $d_{ii}$ of radius $\|E\|_\infty$.

  • $\begingroup$ why are we try to show that the disks will not overlap ? $\endgroup$ – user521337 Jan 11 at 8:31
  • $\begingroup$ Because if the disks have no overlap, then each contains one eigenvalue. $\endgroup$ – Omnomnomnom Jan 11 at 14:55
  • $\begingroup$ Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $\|E\|_\infty$. $\endgroup$ – Omnomnomnom Jan 11 at 17:03

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