Gershgorin Circle Theorem - Statement and Proof

Statement and Proof

Let A be a complex n × n matrix, with entries . For i ∈ {1, …, n} let be the sum of the absolute values of the non-diagonal entries in the ith row. Let D(a_ii, R_i) be the closed disc centered at a_ii with radius R_i. Such a disc is called a Gershgorin disc.

Theorem: Every eigenvalue of A lies within at least one of the Gershgorin discs D(a_ii, R_i).

Proof: Let λ be an eigenvalue of A and let x = (x_j) be a corresponding eigenvector. Let i ∈ {1, …, n} be chosen so that |x_i| = max_j |x_j|. (That is to say, choose i so that x_i is the largest (in absolute value) number in the vector x) Then |x_i| > 0, otherwise x = 0. Since x is an eigenvector, Ax = λx, and thus:

So, splitting the sum, we get

We may then divide both sides by x_i (choosing i as we explained we can be sure that x_i ≠ 0) and take the absolute value to obtain

where the last inequality is valid because

Corollary: The eigenvalues of A must also lie within the Gershgorin discs C_j corresponding to the columns of A.

Proof: Apply the Theorem to AT.

Example For a diagonal matrix, the Gershgorin discs coincide with the spectrum. Conversely, if the Gershgorin discs coincide with the spectrum, the matrix is diagonal.