LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-02 (claude-opus-5)

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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.
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In any vector space $0_F v = 0_V$ , $\lambda 0_V = 0_V$ , $(-\lambda)v = -(\lambda v)$ , $(-1_F)v = -v$ , and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$

Statement

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

$0_F v = 0_V$ ;
$\lambda 0_V = 0_V$ ;
$(-\lambda)v = -(\lambda v)$ , and also $\lambda(-v) = -(\lambda v)$ ;
$(-1_F)v = -v$ ;
if $\lambda v = 0_V$ then $\lambda = 0_F$ or $v = 0_V$ .

Here $0_F$ and $1_F$ are the additive and multiplicative identities of $F$ , $0_V$ is the zero vector, $-\lambda$ is the additive inverse of $\lambda$ 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 $\lambda \in F$ and a vector $v \in V$ .

[L1]

The four scalar axioms: $\lambda(u+w) = \lambda u + \lambda w$ (V2); $(\lambda+\mu)w = \lambda w + \mu w$ (V3); $(\lambda\mu)w = \lambda(\mu w)$ (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: $gx = gy$ or $xg = yg$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $gx = h$ and $xg = h$ each have exactly one solution).

[L4]

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

[L5]

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

Proof

technique · direct

1.1

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

L1L4

1.2

Since $0_V$ is a two-sided identity for $+$ : $0_F v = 0_V + 0_F v$ .

1.3

By (V2) applied to $0_V$ and $0_V$ , and $0_V + 0_V = 0_V$ in $V$ : $\lambda 0_V + \lambda 0_V = \lambda(0_V + 0_V) = \lambda 0_V$ .

L1L2

1.4

Since $0_V$ is a two-sided identity for $+$ : $\lambda 0_V = 0_V + \lambda 0_V$ .

1.5

The vector $\lambda v$ has an additive inverse $-(\lambda v)$ with $\lambda v + (-(\lambda v)) = 0_V$ .

2.1

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.

step 1.1step 1.2L3

2.2

Combining steps 1.3 and 1.4 gives $\lambda 0_V + \lambda 0_V = 0_V + \lambda 0_V$ ; cancelling $\lambda 0_V$ on the right yields $\lambda 0_V = 0_V$ , which is claim 2.

step 1.3step 1.4L3

3.1

By (V3) applied to $\lambda$ and $-\lambda$ , then $\lambda + (-\lambda) = 0_F$ , then claim 1: $\lambda v + (-\lambda)v = (\lambda + (-\lambda))v = 0_F v = 0_V$ .

step 2.1L1L4

3.2

By (V2) applied to $v$ and $-v$ , then $v + (-v) = 0_V$ , then claim 2: $\lambda v + \lambda(-v) = \lambda(v + (-v)) = \lambda 0_V = 0_V$ .

step 2.2L1L2

3.3

Suppose $\lambda v = 0_V$ and $\lambda \ne 0_F$ . Then $\lambda^{-1} \in F$ exists with $\lambda^{-1}\lambda = 1_F$ , so $v = 1_F v = (\lambda^{-1}\lambda)v = \lambda^{-1}(\lambda v) = \lambda^{-1} 0_V = 0_V$ , using (V5), (V4) and claim 2 in turn.

step 2.2L1L4L5

4.1

Steps 3.1 and 1.5 exhibit both $(-\lambda)v$ and $-(\lambda v)$ as vectors $x$ with $\lambda v + x = 0_V$ ; cancelling $\lambda v$ on the left gives $(-\lambda)v = -(\lambda v)$ .

step 3.1step 1.5L3

5.1

Likewise steps 3.2 and 1.5 give $\lambda v + \lambda(-v) = 0_V = \lambda v + (-(\lambda v))$ , and cancelling $\lambda v$ on the left gives $\lambda(-v) = -(\lambda v)$ ; with step 4.1 this is claim 3.

step 3.2step 1.5L3

5.2

Taking $\lambda = 1_F$ in step 4.1 and using (V5): $(-1_F)v = -(1_F v) = -v$ , which is claim 4.

step 4.1L1

6.1

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 $\lambda v = 0_V$ then either $\lambda = 0_F$ , or $\lambda \ne 0_F$ and step 3.3 gives $v = 0_V$ .

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

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 $\operatorname{span}\{v\}$ ( $\operatorname{span}\{v\} = \{\, \lambda v : \lambda \in F \,\}$ , which is $\{0_V\}$ when $v = 0_V$ , and when $v \ne 0_V$ 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: $ab = 0 \Rightarrow a = 0$ 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.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 16 results over 11 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

Vector space (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed., Ch. 1 (standard reference, not scraped)

In any vector space 0Fv=0V0_F v = 0_V0F​v=0V​, λ0V=0V\lambda 0_V = 0_Vλ0V​=0V​, (−λ)v=−(λv)(-\lambda)v = -(\lambda v)(−λ)v=−(λv), (−1F)v=−v(-1_F)v = -v(−1F​)v=−v, and λv=0V\lambda v = 0_Vλv=0V​ forces λ=0F\lambda = 0_Fλ=0F​ or v=0Vv = 0_Vv=0V​