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The Gerschgorin Circle Theorem states that the set of eigenvalues of a matrix $A \in \mathbb{C}^{n \times n}$ is contained in the union of its Gerschgorin discs (i.e., $$\lambda(A) \subseteq \bigcup\limits_{i=1}^{n} G_i$$ for $G_i = \{ z \in \mathbb{C} : \lvert z-a_{ii} \rvert < r_i \}$ and $r_i = \sum_{j \neq i} \lvert a_{ij} \rvert$, and furthermore, that the number of eigenvalues (counted with multiplicity) in each connected component of $\bigcup\limits_{i=1}^{n} G_i$ is equal to the number of $G_i$s in that component.

A basic proof shows that eigenvalues are a continuous function of the entries of the matrix, and so they move continuously throughout each $G_i$.

What I'm confused about is that since the eigenvalues cannot make discrete jumps between $G_i$s, how can we end up with two eigenvalues in a connected component such that the eigenvalues are wholly contained in only one of the two $G_i$s.

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    In your first line, you should replace "eigenvectors" by "eigenvalues". – 2017-01-17
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    Oops, thanks for the correction. – 2017-01-17
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    Not quite sure, but we can imagine that, starting for example with pairwise disjoint discs, and by continuously transform the entries in $A$, we get two of the discs meet ... and then two eigenvalue could be inside both discs. – 2017-01-17
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    Thanks, I see that it's possible in that case for the eigenvalues to be inside the intersection of the two discs, but it's the separate case that I'm confused about (just edited to reflect). That is, suppose two circles $G_i$ and $G_j$ meet, but the eigenvalues are only contained in $G_j$ (which is possible from my reading of the theorem since $G_i \cup G_j$ is a connected component). – 2017-01-17
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    After the two discs have met, and by continuing to transform $A$, the two discs could separate again but leaving the two eigenvalues reside in only one of them. – 2017-01-17
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    !! Great, thank you, I had not thought about this !! Would upvote but am unable to do so. Do you have a formal way of showing this, or would you like to (or want me to) write up this informal explanation as an answer so that I can accept? – 2017-01-17
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    Actually, aren't the radii always increasing if we consider the transformation as $t$ goes from 0 to 1 with $A = D + tF$ where $D$ is diag($A$) and $F$ is the off-diagonal elements? Or is there another proof construction? – 2017-01-17

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Here is an example where one of the $G_i$s doesn't content any eigenvalue.

This is Maple code. The code for the function showgersch is given below.

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with(LinearAlgebra); with(plots); with(plottools);

gersch computes the necessary data

gersch := proc (A)

local n, L;

n := RowDimension(A);

if n <> ColumnDimension(A) then return

else L := [[seq([Re(A[i, i]), Im(A[i, i]), add(abs(A[i, k]), k = 1 .. i-1)+add(abs(A[i, k]), k = i+1 .. n)], i = 1 .. n)], Eigenvalues(A, output = list)];

return evalf(L)

end if

end proc

showgersch display the circles in blue and the eigenvalues as red spots

showgersch := proc (A, xmin, xmax, ymin, ymax)

local L;

L := gersch(A);

display([seq(circle([L[1, k, 1], L[1, k, 2]], L[1, k, 3], color = blue), k = 1 .. nops(L[1]))], [seq(disk([Re(L[2, k]), Im(L[2, k])], 0.5e-1, color = red), k = 1 .. nops(L[2]))], axes = frame, view = [xmin .. xmax, ymin .. ymax])

end proc

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    This is great! I'm curious about `showgersch` -- can you provide a bit of detail? I imagine that you compute $r_i$ and the eigenvalues for the diagonal matrix to get the g-circles? How did you determine that this matrix would work as an example? – 2017-01-19
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    Ok, I will edit my answer to show the Maple code. I didn't have any clue for this counterexample, so I tried a few simple Matrix and this led me to this one. – 2017-01-20