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

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For a finite-dimensional space, $λ$ is an eigenvalue of $T$ if and only if $T - λ I$ is not invertible

Statement

Let $T : V \to V$ be an endomorphism of a finite-dimensional vector space over $F$ , and let $λ \in F$ . Then $λ$ is an eigenvalue of $T$ if and only if $T - λ I_{V}$ is not invertible.

Facts & Assumptions

Given: A finite-dimensional $F$ -vector space $V$ , an endomorphism $T$ of $V$ , and $λ \in F$ .

[L1]

The scalar $λ$ is an eigenvalue of $T$ exactly when $\ker (T - λ I_{V}) \neq {0_{V}}$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

[L2]

A linear map is injective if and only if its kernel is ${0}$ (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial).

[L3]

For a linear map $S : V \to V$ with $V$ finite-dimensional, $\dim V = \dim \ker S + \dim im S$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ). A subspace of $V$ has dimension $\dim V$ if and only if it is all of $V$ (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 2).

[L4]

An invertible linear map has a two-sided linear inverse, and hence is bijective (Invertible linear maps, linear isomorphisms, and inverse linear maps).

Proof

technique · direct

1.1

If $λ$ is an eigenvalue, [L1] and [L2] show that $T - λ I_{V}$ is not injective; it therefore cannot be invertible by [L4].

L1L2L4given

1.2

Conversely, suppose $λ$ is not an eigenvalue. Then [L1] and [L2] make $S = T - λ I_{V}$ injective, so $\ker S = {0}$ . By [L3], $\dim im S = \dim V$ , hence $im S = V$ and $S$ is bijective. Its inverse function is linear: applying $S$ to $S^{- 1} (a u + b v)$ and to $a S^{- 1} (u) + b S^{- 1} (v)$ gives $a u + b v$ in both cases, and injectivity makes the two inputs equal. Thus $S$ is invertible; contraposition gives that noninvertibility forces $λ$ to be an eigenvalue.

L1L2L3L4givenalgebra

2.1

The two implications establish the equivalence. If $V = {0_{V}}$ , [L1] makes the left side false and the unique endomorphism is the identity, so the same argument includes that case.

step 1.1step 1.2L1L4∎

Depends on

Used by

For every finite-dimensional space, σ_F(T) is exactly the set of roots in F of χ_T Theorem

Dependency tree · two levels

34 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, §12.1 (standard reference, not scraped)
M. Khovanov, Linear Algebra II notes, §6 (standard reference, not scraped)

For a finite-dimensional space, λλ is an eigenvalue of TT if and only if T−λIT−λI is not invertible