TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field), let $n \in N$ and let $v : n \to V$ be a finite list (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ).

The span of the image of a list. Whether or not $v$ is injective, $span (v [n]) = {\sum_{i < n} λ_{i} v_{i} : λ : n \to F} .$
Coordinates. $v$ is an ordered basis of $V$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) if and only if for every $x \in V$ there is exactly one $λ : n \to F$ with $x = \sum_{i < n} λ_{i} v_{i}$ . When that holds, this $λ$ is called the coordinate list of $x$ with respect to the ordered basis $v$ , and $λ_{i}$ its $i$ -th coordinate.

The coordinate list is attached to the ordered basis and not to the basis as a set: reordering the list permutes the coordinates of every vector, as the companion page shows on a worked example in $R^{2}$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ , a natural number $n$ and a list $v : n \to V$ .

[L1]

For $T \subseteq V$ , $span (T)$ is a linear subspace of $V$ containing $T$ and contained in every linear subspace of $V$ containing $T$ , and it is exactly the set of linear combinations $\sum_{i < p} μ_{i} y_{i}$ with $y : p \to T$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , $span (S)$ is exactly the set of linear combinations of finite lists of elements of $S$ , and $span (\emptyset) = {0_{V}}$ ).

[L2]

Finite sums: $\sum_{i < 0} u_{i} = 0_{V}$ and $\sum_{i < σ (p)} u_{i} = (\sum_{i < p} u_{i}) + u_{p}$ (The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ); (F1) an all- $0_{V}$ list sums to $0_{V}$ ; (F2) $λ \sum_{i < p} u_{i} + \sum_{i < p} w_{i} = \sum_{i < p} (λ u_{i} + w_{i})$ ; (F3) with (F1), a list vanishing off a single index $j$ sums to its value at $j$ (The sum $U + W$ of two linear subspaces and the sum $\sum_{i < n} U_{i}$ of a finite family).

[L3]

One-step test: a nonempty $W \subseteq V$ with $μ u + w \in W$ for all $μ \in F$ and $u, w \in W$ is a linear subspace of $V$ (One-step subspace test: a nonempty $W \subseteq V$ is a linear subspace if and only if $λ u + v \in W$ for all $λ \in F$ and $u, v \in W$ , Linear subspace of a vector space).

[L4]

The vector space axioms (Vector space over a field) and their consequences (In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ ): $(V, +, 0_{V})$ is an abelian group; (V3) $(λ + μ) y = λ y + μ y$ ; (V4) $(λ μ) y = λ (μ y)$ ; (V5) $1_{F} y = y$ ; $0_{F} y = 0_{V}$ ; and $(- 1_{F}) y = - y$ .

[L5]

An ordered basis of $V$ is an injective list $v : n \to V$ whose image is a basis, equivalently a linearly independent list $v : n \to V$ with $span (v [n]) = V$ ; and a list is linearly independent exactly when it is injective with linearly independent image (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent, Finite sums re-indexed along an injection, with a zero term deleted, and concatenated; and the closure properties of linear independence: an independent list is injective and never $0_{V}$ , its sublists are independent, a list is independent exactly when it is injective with linearly independent image, and every subset of a linearly independent set is linearly independent, claim 6).

[L6]

$F$ is a field: it has $0_{F}$ and $1_{F}$ , and every $μ \in F$ has an additive inverse $- μ$ with $μ + (- μ) = 0_{F}$ (Field).

[L7]

Images and injectivity are as in Injection, surjection, bijection; $n = {0, \dots, n - 1}$ (The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ).

Proof

technique · direct

1.1L2L3L4

Write $W : = {\sum_{i < n} λ_{i} v_{i} : λ : n \to F}$ . It is a linear subspace of $V$ : it contains $0_{V}$ , taking $λ_{i} = 0_{F}$ for every $i$ , since then every entry $0_{F} v_{i}$ is $0_{V}$ and (F1) applies; and for $μ \in F$ and elements $x = \sum_{i < n} λ_{i} v_{i}$ and $y = \sum_{i < n} κ_{i} v_{i}$ of $W$ , the identity (F2) gives $μ x + y = \sum_{i < n} (μ (λ_{i} v_{i}) + κ_{i} v_{i}) = \sum_{i < n} (μ λ_{i} + κ_{i}) v_{i}$ by (V4) and (V3), which again lies in $W$ . So the one-step test applies.

1.2L2L4

$v [n] \subseteq W$ : for $j < n$ take $λ_{j} = 1_{F}$ and $λ_{i} = 0_{F}$ for $i \neq j$ ; the list $i \mapsto λ_{i} v_{i}$ then vanishes off the single index $j$ and has the value $1_{F} v_{j} = v_{j}$ there, so it sums to $v_{j}$ .

1.3L1L7

$W \subseteq span (v [n])$ : each $\sum_{i < n} λ_{i} v_{i}$ is a linear combination of the list $v$ , which takes its values in $v [n]$ , so it lies in the span of $v [n]$ .

2.1step 1.1step 1.2step 1.3L1

Claim 1. By steps 1.1 and 1.2 the set $W$ is a linear subspace of $V$ containing $v [n]$ , so minimality of the span gives $span (v [n]) \subseteq W$ ; with step 1.3 the two sets are equal.

3.1step 2.1L2L4L5L6

Claim 2, from left to right. Let $v$ be an ordered basis, so the list $v$ is linearly independent and $span (v [n]) = V$ . Existence: by step 2.1 every $x \in V = span (v [n])$ lies in $W$ , that is, $x = \sum_{i < n} λ_{i} v_{i}$ for some $λ : n \to F$ . Uniqueness: if $\sum_{i < n} λ_{i} v_{i} = \sum_{i < n} κ_{i} v_{i} = x$ , apply (F2) with the scalar $- 1_{F}$ to the lists $i \mapsto κ_{i} v_{i}$ and $i \mapsto λ_{i} v_{i}$ ; the left-hand side is $- x + x = 0_{V}$ and the right-hand side is $\sum_{i < n} ((- 1_{F}) (κ_{i} v_{i}) + λ_{i} v_{i}) = \sum_{i < n} (λ_{i} + (- κ_{i})) v_{i}$ by (V4), (V3) and $(- 1_{F}) κ_{i} = - κ_{i}$ . Independence of the list $v$ now gives $λ_{i} + (- κ_{i}) = 0_{F}$ , hence $λ_{i} = κ_{i}$ , for every $i < n$ .

3.2step 2.1L2L4L5

Claim 2, from right to left. Suppose every $x \in V$ is $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ$ . Then $V \subseteq W$ , and step 2.1 gives $W = span (v [n]) \subseteq V$ , so $span (v [n]) = V$ . The list $v$ is independent: if $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ , then $λ$ and the all-zero scalar list both represent $0_{V}$ , the latter by (F1) and $0_{F} v_{i} = 0_{V}$ , so uniqueness at $x = 0_{V}$ forces $λ_{i} = 0_{F}$ for every $i < n$ . Being independent, $v$ is injective with linearly independent image, so $v [n]$ is a basis of $V$ and $v$ is an ordered basis.

4.1step 2.1step 3.1step 3.2∎

Claim 1 is step 2.1, and claim 2 is steps 3.1 and 3.2 together.

Remarks

Claim 1 needs no hypothesis on the list. It says that spanning by a finite set can always be computed with one coefficient per listed vector, repetitions and all. It is claim 2 that turns this into a coordinate system, and what it adds is uniqueness, which is exactly independence.
The assignment $x \mapsto λ$ is deliberately left un-named here. It is a bijection $V \to F^{n}$ compatible with the operations, that is a linear isomorphism; but linear maps are the subject of a later page, and naming the map now would be to use a notion this page does not have. What is used below is only the statement above: existence and uniqueness of the coordinate list.
Reordering is not a harmless relabelling. Two ordered bases with the same image assign different coordinate lists to the same vector, so "the coordinates of $x$ in $B$ " is incomplete language when $B$ is a set. The companion page computes the same vector's coordinates in three ordered bases of $R^{2}$ , two of which have the same image.

A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A finite list v:n→Vv:n→V is an ordered basis if and only if every x∈Vx∈V equals ∑i<nλivi∑i<n​λi​vi​ for exactly one λ:n→Fλ:n→F; those scalars are the coordinates of xx in that ordered basis

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis