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Take the following matrix as an example, $$\left( \begin{array}{ccc} -1 & -i & 0 \\ \frac{i}{2} & 0 & 0 \\ 0 & i & 1 \\ \end{array} \right)$$ The eigenvalues are ${-1.37, 1, 0.37}$, and the plot of the Gerschgorin's disks is: enter image description here

We see that each of the disks owns its eigenvalue. Take a second example: $$\left( \begin{array}{ccc} -1 & -i & 0 \\ \frac{i}{2} & 0 & 0 \\ 0 & i & i \\ \end{array} \right)$$

The eigenvalues are ${-1.37, i, 0.37}$, and the plot of the Gerschgorin's disks is: enter image description here

Again we see that each of the disks owns its eigenvalue. Take a third example: $$\left( \begin{array}{ccc} -1 & i & 0 \\ \frac{i}{2} & 0 & 0 \\ 0 & -i & -i \\ \end{array} \right)$$ The eigenvalues are ${ -i, -\frac{1}{2} + \frac{i}{2}, -\frac{1}{2} - \frac{i}{2} }$, and the plot of the Gerschgorin's disks is: enter image description here

We see that one of the disks doesn't own an eigenvalue. My question is how to tell when each disk own its eigenvalue, and under what condition a disk doesn't own its eigenvalue. And any reference for more details about Gerschgorin's disks?

Putterboy
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    One handy fact about the Gerschgorin disks is that each connected collection of disks has the number of eigenvalues equal to the number of disks- if you have two disjoint collections, one of 3 circles and the other of 2, they will have 3 and 2 eigenvalues respectively. This gives a sufficient condition, in your notation, for each to "own" its eigenvalue; they are all pairwise disjoint. – Clinton Bradford Aug 09 '14 at 03:50

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