Vector Product is Zero only if Factor is Zero

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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.


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

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


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 $\mathbf v \ne \mathbf 0$.


Also see