LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-08-02 (claude-opus-5)

How statement and proof provenance work

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.

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 $v = 0_{V}$

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field). For all $λ \in F$ and $v \in V$ :

$0_{F} v = 0_{V}$ ;
$λ 0_{V} = 0_{V}$ ;
$(- λ) v = - (λ v)$ , and also $λ (- v) = - (λ v)$ ;
$(- 1_{F}) v = - v$ ;
if $λ v = 0_{V}$ then $λ = 0_{F}$ or $v = 0_{V}$ .

Here $0_{F}$ and $1_{F}$ are the additive and multiplicative identities of $F$ , $0_{V}$ is the zero vector, $- λ$ is the additive inverse of $λ$ in $F$ , and $- v$ is the additive inverse of $v$ in the abelian group $(V, +, 0_{V})$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ with axioms (V1)–(V5) (Vector space over a field), a scalar $λ \in F$ and a vector $v \in V$ .

[L1]

The four scalar axioms: $λ (u + w) = λ u + λ w$ (V2); $(λ + μ) w = λ w + μ w$ (V3); $(λ μ) w = λ (μ w)$ (V4); $1_{F} w = w$ (V5) (Vector space over a field).

[L2]

$(V, +, 0_{V})$ is an abelian group (V1): addition is associative and commutative, $0_{V}$ is a two-sided identity, and each $w \in V$ has an additive inverse $- w$ with $w + (- w) = 0_{V} = (- w) + w$ (Vector space over a field, Group and abelian group).

[L3]

Cancellation in a group, read additively: if $x + y = x + z$ then $y = z$ , and if $y + x = z + x$ then $y = z$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

[L4]

Field arithmetic (Field): $0_{F} + 0_{F} = 0_{F}$ ; $μ + (- μ) = 0_{F}$ for every $μ \in F$ ; $1_{F}$ is the multiplicative identity; multiplication is associative; and every $μ \neq 0_{F}$ has a multiplicative inverse $μ^{- 1}$ with $μ^{- 1} μ = 1_{F}$ .

[L5]

The identities $0_{F}$ , $1_{F}$ and the inverses $- μ$ , $μ^{- 1}$ of a field are unique, so those notations denote well-defined elements (Identities and inverses in a field are unique).

Proof

technique · direct

1.1L1L4

By (V3) applied to $0_{F}$ and $0_{F}$ , and $0_{F} + 0_{F} = 0_{F}$ in $F$ : $0_{F} v + 0_{F} v = (0_{F} + 0_{F}) v = 0_{F} v$ .

1.2L2

Since $0_{V}$ is a two-sided identity for $+$ : $0_{F} v = 0_{V} + 0_{F} v$ .

1.3L1L2

By (V2) applied to $0_{V}$ and $0_{V}$ , and $0_{V} + 0_{V} = 0_{V}$ in $V$ : $λ 0_{V} + λ 0_{V} = λ (0_{V} + 0_{V}) = λ 0_{V}$ .

1.4L2

Since $0_{V}$ is a two-sided identity for $+$ : $λ 0_{V} = 0_{V} + λ 0_{V}$ .

1.5L2

The vector $λ v$ has an additive inverse $- (λ v)$ with $λ v + (- (λ v)) = 0_{V}$ .

2.1step 1.1step 1.2L3

Combining steps 1.1 and 1.2 gives $0_{F} v + 0_{F} v = 0_{V} + 0_{F} v$ ; cancelling $0_{F} v$ on the right yields $0_{F} v = 0_{V}$ , which is claim 1.

2.2step 1.3step 1.4L3

Combining steps 1.3 and 1.4 gives $λ 0_{V} + λ 0_{V} = 0_{V} + λ 0_{V}$ ; cancelling $λ 0_{V}$ on the right yields $λ 0_{V} = 0_{V}$ , which is claim 2.

3.1step 2.1L1L4

By (V3) applied to $λ$ and $- λ$ , then $λ + (- λ) = 0_{F}$ , then claim 1: $λ v + (- λ) v = (λ + (- λ)) v = 0_{F} v = 0_{V}$ .

3.2step 2.2L1L2

By (V2) applied to $v$ and $- v$ , then $v + (- v) = 0_{V}$ , then claim 2: $λ v + λ (- v) = λ (v + (- v)) = λ 0_{V} = 0_{V}$ .

3.3step 2.2L1L4L5

Suppose $λ v = 0_{V}$ and $λ \neq 0_{F}$ . Then $λ^{- 1} \in F$ exists with $λ^{- 1} λ = 1_{F}$ , so $v = 1_{F} v = (λ^{- 1} λ) v = λ^{- 1} (λ v) = λ^{- 1} 0_{V} = 0_{V}$ , using (V5), (V4) and claim 2 in turn.

4.1step 3.1step 1.5L3

Steps 3.1 and 1.5 exhibit both $(- λ) v$ and $- (λ v)$ as vectors $x$ with $λ v + x = 0_{V}$ ; cancelling $λ v$ on the left gives $(- λ) v = - (λ v)$ .

5.1step 3.2step 1.5L3

Likewise steps 3.2 and 1.5 give $λ v + λ (- v) = 0_{V} = λ v + (- (λ v))$ , and cancelling $λ v$ on the left gives $λ (- v) = - (λ v)$ ; with step 4.1 this is claim 3.

5.2step 4.1L1

Taking $λ = 1_{F}$ in step 4.1 and using (V5): $(- 1_{F}) v = - (1_{F} v) = - v$ , which is claim 4.

6.1step 2.1step 2.2step 3.3step 4.1step 5.1step 5.2∎

Claim 1 is step 2.1, claim 2 is step 2.2, claim 3 is steps 4.1 and 5.1, and claim 4 is step 5.2; for claim 5, if $λ v = 0_{V}$ then either $λ = 0_{F}$ , or $λ \neq 0_{F}$ and step 3.3 gives $v = 0_{V}$ .

Remarks

None of this is an axiom. The scalar axioms (V2)–(V5) say nothing directly about $0_{F}$ , $0_{V}$ or negatives; each claim above is extracted by writing one element in two ways and cancelling in the abelian group $(V, +, 0_{V})$ . That is the same device that gives $0 \cdot a = 0$ in a field (Multiplication by zero: $0 \cdot a = 0$ ), and the proofs are deliberately parallel.
Claim 5 is what makes ${0_{V}}$ the only "degenerate" scalar multiple. It is used below to compute $span {v}$ ( $span {v} = {λ v : λ \in F}$ , which is ${0_{V}}$ when $v = 0_{V}$ , and when $v \neq 0_{V}$ contains $0_{V}$ only as the multiple $0_{F} v$ ) and, in that form, is the vector-space analogue of a field having no zero divisors (A field has no zero divisors: $a b = 0 \Rightarrow a = 0$ or $b = 0$ ). Its converse directions, claims 1 and 2, say that both degenerate products really are $0_{V}$ .
Claim 4 is the bridge to the additive group. It says the additive inverse of a vector is a scalar multiple of it, which is why closure under scalar multiplication already forces closure under negatives; that is what makes a linear subspace a subgroup of $(V, +)$ without a separate axiom.

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 $v = 0_{V}$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

In any vector space 0Fv=0V0F​v=0V​, λ0V=0Vλ0V​=0V​, (−λ)v=−(λv)(−λ)v=−(λv), (−1F)v=−v(−1F​)v=−v, and λv=0Vλv=0V​ forces λ=0Fλ=0F​ or v=0Vv=0V​

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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 $v = 0_{V}$