We all know Gershgorin's Circle Theorem, which I will summarise for convenience. Let $A=(a_{ij})$ be an $n\times n$ complex matrix. Define the disks $D_1,\ldots,D_n$ by $$D_i = \Bigl\{ z : |z-a_{ii}|\le \sum_{j\ne i} |a_{ij}|\Bigr\}.$$ Then each eigenvalue of $A$ lies in one of the disks. Moreover, if a connected component of the union of the disks contains $k$ disks, then exactly $k$ eigenvalues of $A$ lie in that union.

My question is when a stronger statement is true. When is it possible to list the eignvalues $\lambda_1,\ldots,\lambda_n$ in such an order that $\lambda_i\in D_i$ for all $i$?

What is a small counterexample for general matrices? Is there a counterexample for real symmetric matrices? Is there a nice family of matrices for which there is no counterexample?

Note that by Hall's marriage theorem, the stronger statement is equivalent to saying that for each $k$, the union of any $k$ disks includes at least $k$ eigenvalues.

connected componentof the union consisting of k disks has exactly k zeroes in it :). – fedja Aug 13 '11 at 16:28