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I am studying Gershgorin’s Circle Theorem:

Let $A$ be a matrix with distinct real eigenvalues. For every diagonal element $a_{ii}$, let $$r_i = \sum_{j \ne i} |a_{ij}|.$$

Let $D_i = \{x: |x – a_{ii}| \le r_i\}$.

Then all eigenvalues lie within the set $\cup D_i$, and so lie somewhere within these sets $D_i$.

However, in my calculations I obtained 4 roots where two of the roots are repeated roots and the other two are distinct roots. It means that I obtained 4 eigenvalues, two repeated and two not repeated.

Can I apply the Gershgorin’s Circle Theorem in this case, where two eigenvalues are not distinct?

Thank you.

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    The Theorem states that for all square matrices $A$ over $\mathbb{C}$, every eigenvalue of $A$ lies in at least one of the discs $D_i$. Hence all eigenvalues lie in the union of the discs. It does not matter if $A$ has repeated eigenvalues of not. – 2017-02-19
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    The "basic" Gerschgorin (which doesn't say anything about how the disks are populated) doesn't care about this at all. All it says is that all eigenvalues fall into the union of the disks. There is a technicality in the "refined" Gerschgorin (the one that says something about how the disks are populated). In this case repeated eigenvalues have to be counted according to their multiplicity. – 2017-02-19

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Yes. If you study the proof, you will find that the eigenvalues need not be distinct to reach the theorem's conclusion.

In fact, your statement of the theorem's hypothesis is overly restrictive: the eigenvalues may be repeated or complex.