TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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Complexification preserves the characteristic and minimal polynomials of a finite-dimensional real operator

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional real vector space and let $T_{C}$ be its complexification. Then

$χ_{T_{C}} = χ_{T}, μ_{T_{C}} = μ_{T},$

where the minimal polynomial of $T$ is regarded as an element of $R [x] \subseteq C [x]$ .

Facts & Assumptions

Given: An endomorphism $T$ of a finite-dimensional real vector space $V$ .

[L1]

A real ordered basis becomes an ordered complex basis after complexification (A real basis becomes a complex basis after complexification, so $\dim_{C} (C \otimes_{R} V) = \dim_{R} V$ ).

[L2]

The complexification of a map is $T_{C} (z \otimes v) = z \otimes T (v)$ (Complexification of a real-linear map).

[L3]

The characteristic polynomial of an operator is the characteristic polynomial of its matrix in any ordered basis, independent of the choice (The basis-independent characteristic polynomial $χ_{T}$ of an endomorphism of a finite-dimensional space, including $χ_{T} = 1$ in dimension zero).

[L4]

An operator and its matrix in an ordered basis have the same minimal polynomial (The minimal polynomial is unchanged by choosing a matrix representation or replacing a matrix by a similar one).

[L5]

For a field extension $K / F$ , a matrix $A \in M_{n} (F)$ has the same minimal polynomial over $F$ and over $K$ (For a matrix over a field, extending the scalar field does not change its minimal polynomial).

Proof

technique · direct

1.1

Choose an ordered real basis $B = (v_{1}, \dots, v_{n})$ of $V$ ; by [L1], $(ι v_{1}, \dots, ι v_{n})$ is an ordered complex basis of $V_{C}$ , and by [L2] the matrix of $T_{C}$ in it equals the matrix $A = [T]_{B}$ , because $T_{C} (ι v_{j}) = ι T v_{j}$ .

L1L2

2.1

By [L3], $χ_{T}$ is the characteristic polynomial of $A$ over $R$ and $χ_{T_{C}}$ is the characteristic polynomial of the same matrix $A$ over $C$ ; the two polynomials have the same coefficients, so $χ_{T_{C}} = χ_{T}$ .

step 1.1L3

2.2

By [L4], $μ_{T}$ is the minimal polynomial of $A$ over $R$ and $μ_{T_{C}}$ is the minimal polynomial of $A$ over $C$ .

step 1.1L4

3.1

By [L5] with $K = C$ and $F = R$ , the minimal polynomial of $A$ is the same element of $R [x] \subseteq C [x]$ in both readings, so step 2.2 gives $μ_{T_{C}} = μ_{T}$ .

step 2.2L5

4.1

Steps 2.1 and 3.1 prove both claimed equalities.

step 2.1step 3.1∎

Depends on

Used by

The real quarter-turn diagonalises after complexification but has no real eigenvector Example

Dependency tree · two levels

21 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

Keith Conrad, Complexification (notes) (standard reference, not scraped)