TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Over every extension field, a scalar is an eigenvalue of the extended matrix exactly when it is a root of the minimal polynomial

Statement

Let $A \in M_{n} (F)$ , let $K / F$ be a field extension, and let $λ \in K$ . Then $λ$ is an eigenvalue of the matrix $A$ acting on $K^{n}$ if and only if

$μ_{A} (λ) = 0,$

where $μ_{A} \in F [x]$ is the minimal polynomial over $F$ . For $n = 0$ , both sets are empty.

Facts & Assumptions

Given: A field extension $K / F$ , a matrix $A \in M_{n} (F)$ , and $λ \in K$ .

[L1]

Extending the scalar field does not change the minimal polynomial of $A$ (For a matrix over a field, extending the scalar field does not change its minimal polynomial).

[L2]

The minimal polynomial divides the characteristic polynomial (The minimal polynomial divides the characteristic polynomial, $μ_{T} ∣ χ_{T}$ ).

[L3]

Over any field, a scalar is an eigenvalue exactly when it is a root of the characteristic polynomial (For every finite-dimensional space, $σ_{F} (T)$ is exactly the set of roots in $F$ of $χ_{T}$ ).

[L4]

Polynomial evaluation is $p (A) = \sum a_{k} A^{k}$ (Polynomial evaluation at an endomorphism: $p (T) = \sum_{k} a_{k} T^{k}$ ).

Proof

technique · direct

1.1

Suppose $A v = λ v$ for some nonzero $v \in K^{n}$ . Induction gives $A^{k} v = λ^{k} v$ , so [L4] gives $p (A) v = p (λ) v$ for every $p \in K [x]$ . Taking $p = μ_{A}$ and using [L1] yields $0 = μ_{A} (A) v = μ_{A} (λ) v$ , hence $μ_{A} (λ) = 0$ .

L1L4algebra

1.2

Conversely, if $μ_{A} (λ) = 0$ , then [L2] gives $χ_{A} (λ) = 0$ . The determinant formula for $x I - A$ is unchanged after embedding $F$ in $K$ , so [L3] applied over $K$ says $λ$ is an eigenvalue of $A$ on $K^{n}$ .

L2L3

2.1

When $n = 0$ , $μ_{A} = 1$ by [L1], so it has no roots, while the zero space has no nonzero eigenvector.

L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 12 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

Keith Conrad, The Minimal Polynomial and Some Applications, Theorem 4.7 (standard reference, not scraped)
Keith Conrad, Potential Diagonalizability, Theorem 4(1) (standard reference, not scraped)