In order to estimate the eigenvalues of a real symmetric $n\times n$ matrix, I intend to use the Gershgorin Circle Theorem. Unfortunately, the examples one might find on the internet are a bit confusing; What would be the mathematical formula for deriving the eigenvalue estimates?

I understand that certain disks are formed, each centered at the diagonal entry, with the radius equal to the summation of absolute values of the associated off-diagonal row entries. (The example from


is clear with the disks, but not with the eigenvalue estimation) Which steps to take from this point to get the estimates on the eigenvalues?

From Theorem 2.1 in


one could understand the eigenvalue $ranges$, but the example 2.3 from the above paper gives concrete eigenvalue estimates (some of which are negative). I would appreciate if someone explains this.


1 Answer 1


As far as I understand, Gerschgorin's theorem does not tell you anything about the eigenvalues themselves (say, their exact values, their distribution, etc). It only tells us that each one of the eigenvalues is contained in at least one of the Gerschgorin's discs. In particular, let

$$R = \max_{1\leq i\leq n}\{R_i\},$$

where $\{R_i\}_{1\leq i\leq n}$ are the radii of Gerschgorin's discs. Let

$$A = \max_{1\leq i \leq n}\{a_{ii}\},$$

where $\{a_{ii}\}_{1\leq i\leq n}$ are the diagonal entries of the matrix. Then each eigenvalue of the given matrix lies inside the disc of radius $A + R$ centered at the origin. In particular, no eigenvalue of the given matrix can exceed $A + R$ in magnitude.

Moreover, as far as I understand from the theorem, it isn't necessarily true that there is at least one eigenvalue in each of the Gerschgorin's discs (of course this cannot be true).

  • $\begingroup$ Old post, but to slightly dispute/clarify: The theorem you provide is more of a corollary of Gerschgorin's theorem rather than the result itself. Gerschgorin says all eigenvalues lie in union of the disks and if the disks are all disjoint then each has its own eigenvalue. By putting the union into a disk around the origin you lose almost all information about the eigenvalues' locations. Two corollaries off the top of my head: 1) If $|a_{ii}| > R_i$ for all $i$ then zero isn't in any disk, so the matrix is invertible. 2) If a_{ii} > R_i for all i then a_{ii} is Positive Definite. $\endgroup$
    – adfriedman
    Jul 25, 2023 at 21:46

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