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In short, following a question from my students, I am trying to find a special case where all the eigenvalues of a matrix lie within only one circle, but not in the others, and the other circles are not completely contained in it.

Reminder: Gershgorin circle theorem

Given a matrix $A\in\mathbb{C}^{n\times n}$, define the disks $D_1,\ldots,D_n$ as follows: $$D_i = \Bigl\{ z : |z-a_{ii}|\le \sum_{j\ne i} |a_{ij}|\Bigr\}.$$ The theorem states that all eigenvalues of $A$ lie within (at least) 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.

I am trying to find an example where all eigenvalues lie within only one circle (of course, all circles must partially overlap for this).

For $n=2$

Such an example is easy to find. For instance, $ A=\begin{bmatrix} 7 & 9 \\ -5 & -5 \end{bmatrix} $ has two eigenvalues $\left\{1+3j, 1-3j\right\}$, which are only inside one Gershgorin circle, as illustrated below:

2x2 example

For $n\ge 3$

I could find many examples (by simulation) where one circle completely contains all other circles:

3x3 example

But I am looking for an example where the circles overlap, but are not all contained inside one of the circles. Like in following (hand made) graphs:

Interesting case

I already randomized billions of such graphs without any success, so I am beginning wonder:

Is there an interesting mathematical property making this case impossible?

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    $\begingroup$ I guess a related question would be: can a Gershgorin circle not intersect with any other circle that contains an eigenvalue. $\endgroup$
    – Itay
    Jan 22, 2020 at 6:34

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