# Disjoint Gershgorin disks $\Rightarrow$ each contains exactly one eigenvalue

It is an exercise in Peter Lax's book Linear Algebra that if all the Gershgorin disks

$$D_i := \{z\in \mathbb{C} : |a_{ii} - z| \leq \sum_{i \neq j} |a_{ij}|\}$$

are disjoint, then each disk must contain exactly one eigenvalue of the matrix $A = (a_{ij})$.

What I've tried so far: The proof of Gershgorin's theorem involves selecting the largest-modulus entry $v_i$ of an eigenvector $v$, and then showing that the corresponding eigenvalue $\lambda$ must be in $D_i$.

It would suffice if we could show that for each $i \in \{1,2,\dotsc, n\}$ there was an eigenvector $v$ such that $\max_{j}|v_j| = i$. But I'm not sure how we could show this...

Let $D$ be the diagonal part of $A$ and $M=A-D$ so that $A=D+M$. Let $\lambda_i(s)$ be the $i$th eigenvalue, counting multiplicities, of $D+sM$.
When $s=1$, we have our original case. When $s=0$, we have a diagonal matrix. As $s$ increases from $0$ to $1$, we are blowing up the radii of the disks given by Gershgorin's theorem.
Keeping this in mind, and noting that $\lambda_i(s)$ is continuous (implicit function theorem), consider that one of the disjoint disks $D_i$ does not have an eigenvalue. Then think about the situation as $s \to 0$. You can work out the contradiction--it's great to think about geometrically.
• Can you elaborate a bit how the implicit function theorem implies the continuity of $\lambda_i(s)$? – Eric Auld Aug 25 '13 at 17:46
• Okay, so in reality, you probably have to be careful of double roots, but otherwise you look at the characteristic equation for the eigenvalues and note that each zero, the appropriate derivative will be non-zero if it is a single root. There should be plenty online about the continuity of eigenvalues under sequences of matrices. Perhaps you can prove that if the original $\lambda$s are unique, then so will $\lambda(s)$. But you are right, this requires more consideration. – abnry Aug 25 '13 at 19:20