04-25-2025

Gershgorin and his circles

A theorem I love from linear algebra is the Gershgorin circle theorem which relates the spectrum of a $n \times n$ complex matrix to $n$ disks in the complex plane. Given $A \in \mathbb{C}^{n \times n}$ with spectrum A complex number $\lambda \in \mathbb{C}$ is an eigenvalue of $A$ if and only if there exists $v \in \mathbb{C}^{n}$ with $v \neq 0$ such that $$ Av = \lambda v $$or equivalently, $(A - \lambda I_{n})v = 0$. This implies that $A - \lambda I_{n}$ is a singular matrix. $$ \sigma(A) = \{ \lambda \in \mathbb{C} \ : \ \det(A - \lambda I_{n}) = 0 \} $$ we define $r_{i}$ to be the absolute row sum of the $i$-th row of $A$ with the diagonal entry $a_{ii}$ deleted, i.e. $$ r_{i} := \sum_{j \neq i} \lvert a_{ij} \rvert $$ The closed disk in the complex plane $$ D_{i} = \{ z \in \mathbb{C} \ : \ \lvert z-a_{ii} \rvert \leq r_{i} \} $$ with center $a_{ii}$ and radius $r_{i}$ is the $i$-th Gershgorin disk. The original result of Gershgorin states that the spectrum of $A$ is completely contained in the collection of all such disks. This collection is called the Gershgorin set which is closed and bounded in $\mathbb{C}$.

Theorem (Gershgorin circle theorem)

Let $A = [a_{ij}]_{i,j=1,\dots,n} \in \mathbb{C}^{n \times n}$ be a complex square matrix with spectrum $\sigma(A)$. Then, for every eigenvalue $\lambda \in \sigma(A)$ there exists an index $k \in \{1,\dots, n\}$ such that $$ \lvert \lambda - a_{kk} \rvert \leq r_{k}(A) $$ i.e., $\lambda$ lies in the Gershgorin disk $D_{k}(A)$. Consequently, $$ \sigma(A) \subseteq \bigcup_{i=1}^{\ \ n} D_{i}(A) $$

Show Proof.

Let $\lambda$ be an eigenvalue of $A$ with eigenvector $v \in \mathbb{C}^n$. Since $v \neq 0$ we must have $$ 0 < \lvert v_{k} \rvert = \max\{ \lvert v_{i} \rvert \ : \ i=1,\dots,n \} = \lVert v \rVert_{1} $$ for some $k \in \{1,\dots,n \}$. Using index notation we can write $$ \sum_{j=1}^{\ \ \ n} a_{kj} v_{j} = \lambda v_{k} $$ or equivalently, $$ (\lambda - a_{kk}) v_{k} = \sum_{j \neq k} a_{kj} v_{j} $$ Taking the absolute values in the above equation and using the triangle inequality we get $$ \left\lvert \lambda - a_{kk} \right\rvert \lvert v_{k} \rvert = \left\lvert \sum_{j \neq k} a_{kj} v_{j} \right\rvert \leq \sum_{j \neq k} \lvert a_{kj} v_{j} \rvert \leq \lvert v_{k} \rvert \sum_{j \neq k} \lvert a_{kj} \rvert = \lvert v_{k} \rvert r_{k}(A) $$ Dividing by $\lvert v_{k} \rvert$ gives $\lvert \lambda - a_{kk} \rvert \leq r_{k}(A)$.

If we construct a family of matrices where the diagonal entries (which determine the centers of the circles) and the off-diagonal entries (which determine the radii) vary smoothly, we can use Gershgorin’s theorem to create some pretty plots!

Below is a plot of the Gershgorin sets of the one-parameter family of matrices $A_{t}$ with diagonal elements The $k$-th diagonal entry traces a unit circle in the complex plane $t \mapsto e^{2k\pi it}$. $$ a_{kk} = e^{2k\pi it} $$ and off-diagonal elements that sum to one. This ensures that all circles have the same size and overlap. You can use the slider below to vary $t \in [0, 1]$.