Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.