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

1.1L1L2L4given

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

1.2L1L2L3L4givenalgebra

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.

2.1step 1.1step 1.2L1L4∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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