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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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Every eigenvalue lies in some Gershgorin disk

Statement

Let T:VV 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 Di(T,B).

Facts & Assumptions

Given: A finite-dimensional complex vector space V, a linear endomorphism T:VV, an ordered basis B=(b1,,bn), and the matrix [T]BB=(aij).

[L2]

The columns of [v]B are the coordinates of v in the basis B (Coordinate columns [v]B and matrices [T]BC of linear maps relative to ordered bases).

Proof

technique · direct
1.1

Let λ be an eigenvalue. By [L1], choose a nonzero eigenvector v, and write its coordinate column as [v]B=(x1,,xn)T. Choose k with xk=maxjxj>0.

L1L2choose
2.1

The k-th row of the equation [T]BB[v]B=λ[v]B is jakjxj=λxk, so (λakk)xk=jkakjxj. Taking absolute values and using the maximality of xk yields λakkxkjkakjxj(jkakj)xk. Because xk0, dividing by xk gives λakkjkakj, so λDk(T,B).

step 1.1algebra

Depends on

Used by

Dependency tree · two levels

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Sources