Gershgorin's Circle Theorem: Trapping Eigenvalues in Discs

The precise statement

Let A = (aᵢⱼ) be an n×n matrix over ℂ. For each row index i ∈ {1, …, n} define the deleted absolute row sum

• Rᵢ = ∑ⱼ≠ᵢ |aᵢⱼ|,

and the Gershgorin disc Dᵢ = { z ∈ ℂ : |z − aᵢᵢ| ≤ Rᵢ }, the closed disc centered at the diagonal entry aᵢᵢ with radius Rᵢ. The theorem asserts:

• (Localization) Every eigenvalue λ of A satisfies λ ∈ ⋃ᵢ₌₁ⁿ Dᵢ.
• (Counting refinement) If a union of k of the discs is disjoint from the other n − k, then that union contains exactly k eigenvalues, counted with algebraic multiplicity.

The only hypothesis is that A is a finite square complex matrix — no symmetry, no normality, no diagonalizability. Because A and its transpose Aᵀ share the same eigenvalues, the identical statement holds with column sums Cⱼ = ∑ᵢ≠ⱼ |aᵢⱼ|, so the spectrum also lies in the union of column discs. Intersecting the row-disc region with the column-disc region can only help.

The picture and the intuition

Picture the diagonal entries a₁₁, …, aₙₙ as n pins dropped into the complex plane. If A were diagonal, those pins are the eigenvalues — each disc has radius zero. The off-diagonal entries are the coupling between coordinates, the mess that pushes eigenvalues away from the diagonal. Gershgorin says the push is bounded: an eigenvalue can wander away from aᵢᵢ by at most the total off-diagonal weight in that row.

So a matrix that is diagonally dominant — each |aᵢᵢ| strictly larger than its Rᵢ — has discs that cannot reach the origin, which is exactly why such matrices are invertible (0 is not an eigenvalue). Nearly-diagonal matrices give tiny discs and hence excellent eigenvalue estimates for free.

The geometry is honest about its own limits: the region is a union, not an intersection. Big off-diagonal entries make fat discs that may overlap and merge into one blob, and then the theorem tells you little more than 'the eigenvalues are somewhere in this large region.' The art is to first rescale the matrix so the discs shrink.

The key idea of the proof

The whole proof is one clever choice. Let λ be an eigenvalue with eigenvector x ≠ 0, so Ax = λx. Pick the index i where the coordinate is largest in modulus: |xᵢ| ≥ |xⱼ| for all j, and note xᵢ ≠ 0. Read off row i of the equation Ax = λx:

• ∑ⱼ aᵢⱼ xⱼ = λ xᵢ.

Move the diagonal term to the right and isolate everything else:

• (λ − aᵢᵢ) xᵢ = ∑ⱼ≠ᵢ aᵢⱼ xⱼ.

Take absolute values and divide by |xᵢ| > 0. By the triangle inequality and the maximality of |xᵢ| (so every |xⱼ|/|xᵢ| ≤ 1):

• |λ − aᵢᵢ| = |∑ⱼ≠ᵢ aᵢⱼ xⱼ / xᵢ| ≤ ∑ⱼ≠ᵢ |aᵢⱼ| · (|xⱼ|/|xᵢ|) ≤ ∑ⱼ≠ᵢ |aᵢⱼ| = Rᵢ.

Hence λ ∈ Dᵢ. The counting refinement follows by a continuity/homotopy argument: deform A(t) = D + t·(A − D) from the diagonal D to A; the eigenvalues move continuously in t, discs grow continuously, and eigenvalues trapped in a component of discs cannot jump across the gap to another disjoint component.

A worked example

Take the real matrix

• A = [[10, 1, 0], [1, 2, 1], [−1, 0, 3]].

Row discs: center 10 radius |1|+|0| = 1, so D₁ = D(10, 1); center 2 radius |1|+|1| = 2, so D₂ = D(2, 2); center 3 radius |−1|+|0| = 1, so D₃ = D(3, 1). Every eigenvalue lies in D₁ ∪ D₂ ∪ D₃.

Now the payoff of disjointness. D₁ = D(10,1) covers the real segment [9, 11]; the other two cover [0, 4]. These are disjoint, so D₁ contains exactly one eigenvalue, real (since a lone complex eigenvalue of a real matrix would need its conjugate in the same disc), sitting in [9, 11]. The true eigenvalues are approximately 10.106, 3.115, and 1.779 — the isolated disc D(10,1) pinned the dominant eigenvalue (≈10.106) to [9, 11] instantly.

Column discs give center 10 radius |1|+|−1| = 2, center 2 radius 1, center 3 radius 1; intersecting row and column pictures sharpens the estimate further. All of this before running any iterative eigensolver.

Why the hypotheses matter — and connections

The finiteness hypothesis is not decorative. On infinite-dimensional spaces the naive 'union of discs' can miss the spectrum entirely: the right shift on ℓ² has all diagonal entries 0 and all row sums 1, so every Gershgorin disc is the closed unit disc D(0,1) — yet its spectrum is the full closed unit disc while its point spectrum is empty. The theorem's finite counting refinement (algebraic multiplicities, characteristic polynomial degree n) has no operator-theoretic analogue there.

Even in finite dimensions, note the discs are closed and the region need not be sharp: an eigenvalue may sit deep inside a fat disc, and the boundary is attained only in special cases (e.g. when the maximal-coordinate inequalities are all equalities). Dropping diagonal dominance loses the invertibility corollary — a singular matrix can have a disc touching 0.

Gershgorin connects to the Perron–Frobenius world (row sums bound the spectral radius of nonnegative matrices), to Levy–Desplanques (strictly diagonally dominant ⇒ nonsingular), and is refined by Brauer's ovals of Cassini (1947), which are always contained in the Gershgorin region.

Applications and significance

Gershgorin is the workhorse of quick spectral bookkeeping:

• Invertibility / conditioning: strict diagonal dominance ⇒ 0 ∉ any disc ⇒ A is nonsingular. This underlies convergence guarantees for Jacobi and Gauss–Seidel iterations.
• Spectral radius bounds: ρ(A) ≤ maxᵢ (|aᵢᵢ| + Rᵢ) ≤ ‖A‖∞, giving stability estimates for dynamical systems and Markov chains without eigenvalue computation.
• Positive definiteness: a real symmetric matrix whose discs all lie in the right half-plane (aᵢᵢ > Rᵢ) is positive definite — a cheap sufficient test used in optimization and finite-element analysis.
• Perturbation and clustering: after a similarity scaling D⁻¹AD, tight discs certify that a computed eigenvalue is genuinely isolated, or that a cluster of k eigenvalues is trapped in a small region.

Its enduring appeal is the ratio of insight to cost: for O(n²) work you get a mathematically guaranteed region — not a heuristic — which is exactly what you want when certifying that an iterative solver has converged or that a discretized operator is stable.

Gershgorin and its close relatives: what each buys you

Result	Localization region	Sharpness / extra power
Gershgorin (row) discs	∪ᵢ D(aᵢᵢ, Rᵢ), Rᵢ = ∑ⱼ≠ᵢ|aᵢⱼ|	O(n²) cost; disjoint disc ⇒ contains exactly one eigenvalue
Gershgorin (column) discs	∪ⱼ D(aⱼⱼ, Cⱼ), Cⱼ = ∑ᵢ≠ⱼ|aᵢⱼ|	Same spectrum since A and Aᵀ share eigenvalues; intersect both
Brauer's ovals of Cassini	∪ᵢ≠ⱼ {z : |z−aᵢᵢ||z−aⱼⱼ| ≤ RᵢRⱼ}	Always ⊆ Gershgorin region; strictly tighter, n(n−1)/2 ovals
Similarity-scaled Gershgorin	Apply to D⁻¹AD (D diagonal, positive)	Shrinks radii adaptively; can separate a cluster of discs
Bauer–Fike (perturbation)	|λ − μ| ≤ κ(V)·‖E‖ for diagonalizable A	Bounds how eigenvalues move under a perturbation E, not localization from scratch