CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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An eigenvalue of algebraic multiplicity one has a one-dimensional eigenspace

Statement

If $λ$ is an eigenvalue of a finite-dimensional endomorphism and has algebraic multiplicity $1$ , then $\dim E_{λ} (T) = 1$ .

Facts & Assumptions

Given: An eigenvalue $λ$ of $T$ whose algebraic multiplicity is $1$ .

[L1]

The geometric multiplicity is $\dim E_{λ} (T)$ , and an eigenvalue has a nonzero eigenspace (Algebraic multiplicity as the exponent of $x - λ$ in $χ_{T}$ , and geometric multiplicity as $\dim E_{λ} (T)$ ).

[L2]

Geometric multiplicity is at most algebraic multiplicity (The geometric multiplicity of an eigenvalue does not exceed its algebraic multiplicity).

Proof

technique · direct

1.1

Since $E_{λ} (T)$ contains an eigenvector, it is nonzero and its finite dimension is at least $1$ .

L1given

1.2

By [L2] and the given algebraic multiplicity, $\dim E_{λ} (T) \leq 1$ .

L1L2given

2.1

Combining steps 1.1 and 1.2 gives $\dim E_{λ} (T) = 1$ .

step 1.1step 1.2algebra∎

Depends on

Used by

Compatible left and right eigenvectors for a simple eigenvalue Definition

Dependency tree · two levels

10 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

H. Pinkham, Linear Algebra, §12.2 (standard reference, not scraped)