Gershgorin Circle Theorem - Strengthening of The Theorem

Strengthening of The Theorem

If one of the discs is disjoint from the others then it contains exactly one eigenvalue. If however it meets another disc it is possible that it contains no eigenvalue (for example, or ). In the general case the theorem can be strengthened as follows:

Theorem: If the union of k discs is disjoint from the union of the other n − k discs then the former union contains exactly k and the latter n − k eigenvalues of A.

Proof: Let D be the diagonal matrix with entries equal to the diagonal entries of A and let

We will use the fact that the eigenvalues are continuous in, and show that if any eigenvalue moves from one of the unions to the other, then it must be outside all the discs for some, which is a contradiction.

The statement is true for . The diagonal entries of are equal to that of A, thus the centers of the Gershgorin circles are the same, however their radii are t times that of A. Therefore the union of the corresponding k discs of is disjoint from the union of the remaining n-k for all t. The discs are closed, so the distance of the two unions for A is . The distance for is a decreasing function of t, so it is always at least d. Since the eigenvalues of are a continuous function of t, for any eigenvalue of in the union the k discs its distance from the union of the other n-k discs is also continuous. Obviously, and assume lies in the union of the n-k discs. Then, so there exists such that . But this means lies outside the Gershgorin discs, which is impossible. Therefore lies in the union of the k discs, and the theorem is proven.