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 $\mathbf v \ne \mathbf 0$.
$\blacksquare$