# On the supremum norm of matrices

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.

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 $$\|E\|_\infty$$.
• Note (as I added to my answer) that each disk fits inside a disk centered at $d_{ii}$ of radius $\|E\|_\infty$. – Omnomnomnom Jan 11 at 17:03