# Vector Product is Zero only if Factor is Zero

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## Theorem

Let $F$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over $F$, as defined by the vector space axioms.

Then:

$\forall \lambda \in F: \forall \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \implies \paren {\lambda = 0_F \lor \mathbf v = \mathbf 0}$

where $\bszero \in \mathbf V$ is the zero vector.

## Proof

Aiming for a contradiction, suppose that:

$\exists \lambda \in F: \exists \mathbf v \in \mathbf V: \lambda \circ \mathbf v = \bszero \land \lambda \ne 0_F \land \mathbf v \ne \bszero$

which is the negation of the exposition of the theorem.

Utilizing the vector space axioms:

 $\ds \lambda \circ \mathbf v$ $=$ $\ds \bszero$ $\ds \leadsto \ \$ $\ds \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}$ $=$ $\ds \lambda^{-1} \circ \mathbf 0$ multiplying both sides by $\lambda^{-1}$ $\ds \leadsto \ \$ $\ds \bszero$ $=$ $\ds \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}$ Zero Vector Scaled is Zero Vector $\ds$ $=$ $\ds \paren {\lambda^{-1} \cdot \lambda} \circ \mathbf v$ $\ds$ $=$ $\ds 1_F \circ \mathbf v$ $\ds$ $=$ $\ds \mathbf v$

which contradicts the assumption that $\mathbf v \ne \mathbf 0$.

$\blacksquare$