Gershgorin circle theorem

Statement

Let $A = (a_{ij})$ be an $n \times n$ matrix over $\C$. For $1 \le i \le n$, let $R_i = \sum_{j \ne i} \abs{a_{ij}}$, and let $D_i = \{z \in \C : \abs{z - a_{ii}} \le R_i\}$ be the closed disc with centre $a_{ii}$ and radius $R_i$. Then each eigenvalue of $A$ lies in at least one of the discs $D_1, \dots, D_n$.

Remarks

The discs are called Gershgorin discs. The proof takes an eigenvector $x$ and an index $i$ with $\abs{x_i}$ maximal, and estimates row $i$ of $Ax = \lambda x$. Since $A$ and $A^T$ have the same eigenvalues, the same holds with column sums. A strictly diagonally dominant matrix, with $\abs{a_{ii}} > R_i$ for all $i$, has no eigenvalue $0$, so it is invertible.

Explanations

L1 human

The eigenvalues of a matrix lie near its diagonal entries, when the other entries are small. Each row gives a disc around its diagonal entry, with the radius equal to the sum of the other absolute values of the row. Every eigenvalue lies in one of these discs.