# Gershgorin's theorem

Complex Plane

$t=$

$M(t)=D+tA=$, $\lambda(t)=$

$M=$

## What's this?

Gershgorin's theorem tells us a region that contains all the eigenvalues of a complex square matrix.

Let $M=(m_{i, j})$ be a complex matrix of order $n$, and $R_i=\sum_{k \ne i} |m_{i, k}|,\quad i=1, 2, \dots, n.$ Let $D_i$ be the closed disc centered at $m_{i, i}$ with radius $R_i$: $D_i=\{\, z \in \mathbb C: |z-m_{i, i}|\le R_i \,\}.$ Then, (i) all the eigenvalues of the matrix $M$ are in $\cup_{i=1}^n D_i$. Furthermore, (ii) if a union of $k$ discs is disjoint from the other $n-k$ discs, then the union contains exactly $k$ eigenvalues.

This program draws the Gershgorin discs (yellow) and the eigenvalues (red) on the complex plane for a given real matrix of order 4. You can observe a continuous deformation used in the proof of the statement (ii) of the theorem and see its correctness.

The proof of (ii) is given as follows. Let $D=\mathop{\mathrm{diag}}(m_{1, 1}, m_{2, 2}, \dots, m_{n, n})$ and $A=M-D$, which are the diagonal part and the non-diagonal part of $M$, respectively. Since $D$ satisfies the statement (ii), the eigenvalues of $M(t)=D+tA$ are continuous in $t$, and the radius $R_i(t)$ increases monotonically and continuously as $t$ varies from $0$ to $1$ continuously, (ii) is true for $M(t)$, $t \in [0, 1]$, from (i). This means that $M=M(1)$ satisfies the statement (ii). The program enables us to understand the above proof in a visual way.

## Libraries

This program uses the following Javascript libraries: