TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every eigenvalue lies in some Gershgorin disk

Statement

Let $T : V \to V$ be a linear endomorphism of a finite-dimensional complex vector space, and let $B$ be an ordered basis of $V$ . Every eigenvalue of $T$ lies in at least one Gershgorin disk $D_{i} (T, B)$ .

Facts & Assumptions

Given: A finite-dimensional complex vector space $V$ , a linear endomorphism $T : V \to V$ , an ordered basis $B = (b_{1}, \dots, b_{n})$ , and the matrix $[T]_{B}^{B} = (a_{i j})$ .

[L1]

An eigenvalue $λ$ has a nonzero eigenvector $v$ with $T v = λ v$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

[L2]

The columns of $[v]_{B}$ are the coordinates of $v$ in the basis $B$ (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

Proof

technique · direct

1.1L1L2choose

Let $λ$ be an eigenvalue. By [L1], choose a nonzero eigenvector $v$ , and write its coordinate column as $[v]_{B} = (x_{1}, \dots, x_{n})^{T}$ . Choose $k$ with $∣ x_{k} ∣ = \max_{j} ∣ x_{j} ∣ > 0$ .

2.1step 1.1algebra∎

The $k$ -th row of the equation $[T]_{B}^{B} [v]_{B} = λ [v]_{B}$ is $\sum_{j} a_{k j} x_{j} = λ x_{k}$ , so $(λ - a_{k k}) x_{k} = \sum_{j \neq k} a_{k j} x_{j}$ . Taking absolute values and using the maximality of $∣ x_{k} ∣$ yields $∣ λ - a_{k k} ∣ ∣ x_{k} ∣ \leq \sum_{j \neq k} ∣ a_{k j} ∣ ∣ x_{j} ∣ \leq (\sum_{j \neq k} ∣ a_{k j} ∣) ∣ x_{k} ∣$ . Because $x_{k} \neq 0$ , dividing by $∣ x_{k} ∣$ gives $∣ λ - a_{k k} ∣ \leq \sum_{j \neq k} ∣ a_{k j} ∣$ , so $λ \in D_{k} (T, B)$ .

Depends on

Used by

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)