Zero Vector Scaled is Zero Vector
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Theorem
Let $\struct {\mathbf V, +, \circ}_K$ be a vector space over a division ring $K$, as defined by the vector space axioms.
Then:
- $\forall \lambda \in K: \lambda \circ \bszero = \bszero$
where $\bszero \in \mathbf V$ is the zero vector.
Proof
\(\ds \lambda \circ \bszero\) | \(=\) | \(\ds \lambda \circ \paren {\bszero + \bszero}\) | Vector Space Axiom $(\text V 3)$: Identity | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \circ \bszero + \lambda \circ \bszero\) | Vector Space Axiom $(\text V 6)$: Distributivity over Vector Addition | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lambda \circ \bszero + \paren {-\lambda \circ \bszero}\) | \(=\) | \(\ds \paren {\lambda \circ \bszero + \lambda \circ \bszero} + \paren {-\lambda \circ \bszero}\) | adding $-\lambda \circ \bszero$ to both sides | ||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \circ \bszero + \paren {\lambda \circ \bszero + \paren {-\lambda \circ \bszero} }\) | Vector Space Axiom $(\text V 2)$: Associativity | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bszero\) | \(=\) | \(\ds \lambda \circ \bszero + \bszero\) | Vector Space Axiom $(\text V 4)$: Inverses | ||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \circ \bszero\) | Vector Space Axiom $(\text V 3)$: Identity |
$\blacksquare$
Also see
- Vector Scaled by Zero is Zero Vector
- Vector Product is Zero only if Factor is Zero
- Zero Vector Space Product iff Factor is Zero
Sources
- 1965: Claude Berge and A. Ghouila-Houri: Programming, Games and Transportation Networks ... (previous) ... (next): $1$. Preliminary ideas; sets, vector spaces: $1.2$. Vector Spaces
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(iv)}$