Zero Vector Space Product iff 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.

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


Then:

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


Proof 1

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:

\(\displaystyle \bszero\) \(=\) \(\displaystyle \lambda^{-1} \circ \bszero\) $\quad$ Zero Vector Scaled is Zero Vector $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}\) $\quad$ as $\lambda \circ \mathbf v = \bszero$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \paren {\lambda^{-1} \circ \lambda} \circ \mathbf v\) $\quad$ Vector Space Axiom $V \, 7$: Associativity with Scalar Multiplication $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 \circ \mathbf v\) $\quad$ Field Axiom $M \, 4$: Inverse Elements for Product $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \mathbf v\) $\quad$ Vector Space Axiom $V \, 8$: Identity for Scalar Multiplication $\quad$

$\blacksquare$


Proof 2

The sufficient condition is proved in Vector Scaled by Zero is Zero Vector, and in Zero Vector Scaled is Zero Vector.

The necessary condition is proved in Vector Product is Zero only if Factor is Zero.

$\blacksquare$


Also see