Zero Vector Space Product iff Factor is Zero
Jump to navigation
Jump to search
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:
| \(\ds \bszero\) | \(=\) | \(\ds \lambda^{-1} \circ \bszero\) | 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 |
$\blacksquare$
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.
$\blacksquare$