TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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The geometric multiplicity of an eigenvalue does not exceed its algebraic multiplicity

Statement

Let $T$ be an endomorphism of a finite-dimensional vector space, and let $λ$ be an eigenvalue. Then

$\dim E_{λ} (T) \leq {mult}_{χ_{T}} (λ) .$

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ , $T \in L (V)$ , and an eigenvalue $λ$ .

[L1]

Geometric multiplicity is $\dim E_{λ} (T)$ , while algebraic multiplicity is the largest exponent of $x - λ$ dividing $χ_{T}$ (Algebraic multiplicity as the exponent of $x - λ$ in $χ_{T}$ , and geometric multiplicity as $\dim E_{λ} (T)$ ).

[L2]

A linearly independent subset of a finite-dimensional space extends, without Choice, to a basis (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , clause 3).

[L3]

The matrix of a linear map records the coordinate columns of the images of the basis vectors (Coordinate columns $[v]_{B}$ and matrices $[T]_{B}^{C}$ of linear maps relative to ordered bases).

[L4]

The characteristic polynomial of a block-triangular matrix is the product of those of its diagonal blocks (The characteristic polynomial of a block upper- or lower-triangular matrix is the product of the characteristic polynomials of its diagonal blocks).

Proof

technique · direct

1.1

Put $g = \dim E_{λ} (T)$ . Choose a basis $e_{1}, \dots, e_{g}$ of the eigenspace and extend it by [L2] to a basis of $V$ .

L1L2givenchoose

2.1

Since $T (e_{i}) = λ e_{i}$ for $i \leq g$ , [L3] makes the matrix of $T$ in this basis block upper triangular with leading diagonal block $λ I_{g}$ .

step 1.1L3given

3.1

By [L4], $χ_{T} (x) = \det (x I_{g} - λ I_{g}) q (x) = (x - λ)^{g} q (x)$ for the characteristic polynomial $q$ of the other diagonal block.

step 2.1L4algebra

4.1

Thus $(x - λ)^{g}$ divides $χ_{T}$ , so the maximal exponent in [L1] is at least $g$ . This is the claimed inequality; $g \geq 1$ because $λ$ is an eigenvalue.

step 3.1L1∎

Depends on

Used by

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

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 98 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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