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

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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.

Finite-dimensional subspaces admit projections without Choice

Statement

Facts & Assumptions

Proof

Sources

Depends on

Dependency tree · 2 levels

Sources