= Gershgorin circle theorem
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{title2=$\lambda\in\bigcup_{i=1}^n D_i(A)$}
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= Gershgorin disc theorem
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The <Gershgorin circle theorem> locates every <eigenvalue> of a <square matrix> using only its entries. A diagonal entry supplies a centre in the <complex plane>, and the sum of the off-diagonal <absolute values> in its row supplies a radius. Small off-diagonal couplings therefore confine the <eigenvalues> near the diagonal entries. This turns a global <eigenvalue> problem into a picture built one row at a time.
Precisely, let $A=(a_{ij})\in\mathbb C^{n\times n}$, with $n\geq1$, and define the closed <Gershgorin discs>
$$
r_i=\sum_{\substack{j=1\\j\ne i}}^n|a_{ij}|,\qquad D_i(A)=\{z\in\mathbb C:|z-a_{ii}|\leq r_i\},\qquad i=1,\ldots,n.
$$
Every <eigenvalue> $\lambda$ of $A$ satisfies $\lambda\in\bigcup_iD_i(A)$, including <eigenvalues> on the boundaries. No symmetry or <diagonalizability> is required. Despite the name, the theorem uses filled <disks>, not just their circular boundaries.https://www.stat.uchicago.edu/~lekheng/courses/302/notes2.pdf[Gene H. Golub, CME 302: Numerical Linear Algebra, Lecture 2, §6, 2005; notes by James Lambers, edited by Lek-Heng Lim]{ref}
The short reason is the <triangle inequality>. For a nonzero <eigenvector> $x$ with $Ax=\lambda x$, choose $i$ with $|x_i|=\max_j|x_j|>0$. The $i$th row gives
$$
(\lambda-a_{ii})x_i=\sum_{j\ne i}a_{ij}x_j,\qquad |\lambda-a_{ii}|\leq\sum_{j\ne i}|a_{ij}|\frac{|x_j|}{|x_i|}\leq r_i.
$$
Thus the largest <eigenvector> coordinate identifies a <Gershgorin disc> containing $\lambda$, even though that <eigenvector> need never be computed to use the bound.
This is useful when a full <eigenvalue> computation would be expensive: forming the <Gershgorin discs> needs one pass over the nonzero entries of a <sparse matrix>. If $0$ lies outside every <Gershgorin disc>, then $A$ is an <invertible matrix>, since zero is not an <eigenvalue>. The condition is exactly $|a_{ii}|>r_i$ in every row, namely <strict diagonal dominance>. This connects <diagonal dominance> to solvability of a <linear system>; the <inverse infinity-norm bound from diagonal dominance> adds a quantitative bound. It is a sufficient test: failure to exclude zero does not prove singularity.
For a concrete <symmetric matrix>, take
$$
A=\begin{pmatrix}2&0.2&0.1\\0.2&5&0.3\\0.1&0.3&9\end{pmatrix},\qquad (r_1,r_2,r_3)=(0.3,0.5,0.4).
$$
The three <Gershgorin discs> have centres $2,5,9$ and are pairwise disjoint. The <finite-dimensional spectral theorem> makes the <eigenvalues> real, so their possible locations reduce to the <intervals> $[1.7,2.3]$, $[4.5,5.5]$, and $[8.6,9.4]$. <Gershgorin eigenvalue counting> gives exactly one <eigenvalue> in each. Independently, the <characteristic polynomial> is
$$
\det(\lambda I-A)=\lambda^3-16\lambda^2+72.86\lambda-89.422,
$$
whose numerical <polynomial roots> are $1.985811$, $4.989942$, and $9.024247$, rounded to six decimal places. All are positive, as <Gershgorin bounds for symmetric matrices> already guarantee without finding those <polynomial roots>.
\Image[/gershgorin-eigenvalue-disks.png]
{title=Three separated Gershgorin discs, each containing one real eigenvalue of the symmetric example}
S. Gerschgorin published the result in 1931 in \i[Über die Abgrenzung der Eigenwerte einer Matrix], in the \i[Bulletin de l'Académie des Sciences de l'URSS], issue 6, pages 749–754. The original paper's §2 states both <eigenvalue> inclusion and counting in separated groups of <Gershgorin discs>; §3 already discusses column bounds and transformations preserving the <eigenvalues>. Thus both the localization picture and ways to sharpen it were part of the original development.https://www.mathnet.ru/php/getFT.phtml?jrnid=im&paperid=5235&what=fullt&option_lang=eng[S. Gerschgorin, Über die Abgrenzung der Eigenwerte einer Matrix, 1931, pp. 749–754; original German paper in the journal archive]{ref}
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