LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Finite-dimensional subspaces admit projections without Choice

Statement

For a subspace $U$ of a finite-dimensional $k$ -vector space $V$ , there is a linear $P : V \to V$ with $P^{2} = P$ , $im P = U$ , and $P ∣_{U} = {id}_{U}$ . No choice axiom is required. This includes $U = 0$ and $U = V$ .

Facts & Assumptions

Given: $V$ finite-dimensional over a field $k$ , and $U \leq V$ .

[F1]

Every independent subset of a subspace of a finite-dimensional space extends to a finite basis, without a choice principle (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 $U = V$ ).

[F2]

Linearity means $P (a v + b w) = a P (v) + b P (w)$ for all scalars and vectors (Linear map between vector spaces over the same field).

Proof

technique · direct

1.1F1

Apply F1 to the empty independent subset of $U$ to obtain a finite basis $u_{1}, \dots, u_{r}$ . This is independent in $V$ , so apply F1 with subspace $V$ to extend it to $u_{1}, \dots, u_{r}, w_{1}, \dots, w_{s}$ . Every vector has an expansion in this basis; two expansions agree coefficientwise because their difference is a zero linear combination of an independent family.

2.1step 1.1F2algebra

Define $P (\sum_{j = 1}^{r} a_{j} u_{j} + \sum_{ℓ = 1}^{s} b_{ℓ} w_{ℓ}) = \sum_{j = 1}^{r} a_{j} u_{j}$ . The uniqueness just proved makes $P$ a well-defined function $V \to V$ . If $v, v^{'}$ have $u$ -coordinates $a_{j}, a_{j}^{'}$ , then $c v + d v^{'}$ has $u$ -coordinates $c a_{j} + d a_{j}^{'}$ . Consequently $P (c v + d v^{'}) = \sum_{j} (c a_{j} + d a_{j}^{'}) u_{j} = c P (v) + d P (v^{'})$ , so $P$ is linear.

3.1step 1.1step 2.1

Each $P (v)$ belongs to $U$ . For $u \in U$ , its basis expression uses only the $u_{j}$ , so $P (u) = u$ . Thus every $u \in U$ is in the image, and $im P = U$ . Since $P (v) \in U$ , also $P (P (v)) = P (v)$ .

4.1F1step 1.1step 2.1step 3.1∎

If $U = 0$ , then $r = 0$ and the formula gives $P = 0$ . If $U = V$ , no added vectors are needed and $P = {id}_{V}$ . If $V = 0$ , both lists are empty and these formulas coincide. Only two applications of the choice-free finite extension result and finite enumerations were used; no simultaneous choice over an infinite family occurs.

Sources

Axler, Linear Algebra Done Right, 4e, 2.32–2.33, pp. 41–42. The local finite-basis supplier works over arbitrary fields, extending Axler’s real/complex convention. Etingof et al., Theorem 4.1.1 proof, p. 62, uses the resulting projection.

Depends on

Used by

A coordinate projection in complex three-space Example
The restriction image on a cobordism boundary is Lagrangian Lemma
Cartan closed subgroup theorem Theorem
Quotient manifold by a closed Lie subgroup Theorem

Dependency tree · two levels

22 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

Etingof et al., Introduction to Representation Theory (standard reference, not scraped)
Sheldon Axler, Linear Algebra Done Right, fourth edition (standard reference, not scraped)