Zero Vector Space Product iff Factor is Zero/Proof 1

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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.

Let $\mathbf v \in \mathbf V, \lambda \in F$.


Then:

$\lambda \circ \mathbf v = \bszero \iff \paren {\lambda = 0_F \lor x = \bszero}$


Proof

A vector space is a module, so all results about modules also apply to vector spaces.

So from Scalar Product with Identity it follows directly that:

$\lambda = 0_F \lor \mathbf v = e \implies \lambda \circ \mathbf v = \bszero$


Next, suppose $\lambda \circ \mathbf v = \bszero$ but $\lambda \ne 0_F$.

Then:

\(\ds \bszero\) \(=\) \(\ds \lambda^{-1} \circ \bszero\) Zero Vector Scaled is Zero Vector
\(\ds \) \(=\) \(\ds \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}\) as $\lambda \circ \mathbf v = \bszero$
\(\ds \) \(=\) \(\ds \paren {\lambda^{-1} \circ \lambda} \circ \mathbf v\) Vector Space Axiom $\text V 7$: Associativity with Scalar Multiplication
\(\ds \) \(=\) \(\ds 1 \circ \mathbf v\) Field Axiom $\text M4$: Inverses for Product
\(\ds \) \(=\) \(\ds \mathbf v\) Vector Space Axiom $\text V 8$: Identity for Scalar Multiplication

$\blacksquare$


Sources

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