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

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An endomorphism of an $n$ -dimensional space has at most $n$ distinct eigenvalues

Statement

If $V$ is an $n$ -dimensional vector space and $T : V \to V$ is linear, then $T$ has at most $n$ distinct eigenvalues.

Facts & Assumptions

Given: An $n$ -dimensional $F$ -vector space $V$ and an endomorphism $T$ .

[L1]

Eigenvectors belonging to pairwise distinct eigenvalues are linearly independent (Eigenvectors belonging to pairwise distinct eigenvalues are linearly independent).

[L2]

Every linearly independent subset of a finite-dimensional vector space is contained, without Choice, in a basis of that space (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).

Proof

technique · direct

1.1

From any finite set of distinct eigenvalues, choose one eigenvector belonging to each; this is a finite sequence of individual choices.

givenchoose

2.1

The chosen vectors are linearly independent by [L1], so [L2] extends them to a basis of $V$ . Every basis of the $n$ -dimensional space has $n$ elements, so the chosen family, and hence the set of chosen eigenvalues, has at most $n$ elements.

step 1.1L1L2algebra

3.1

There cannot be $n + 1$ distinct eigenvalues. Equivalently, $T$ has at most $n$ distinct eigenvalues; when $n = 0$ , no eigenvector exists and the bound is still valid.

step 2.1∎

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: 63 results over 20 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, §12.1 (standard reference, not scraped)

An endomorphism of an nn-dimensional space has at most nn distinct eigenvalues