Zero Vector Scaled is Zero Vector

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Theorem

Let $\struct {\mathbf V, +, \circ}_F$ be a vector space over a field $F$, as defined by the vector space axioms.


Then:

$\forall \lambda \in \mathbb F: \lambda \circ \bszero = \bszero$

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


Sources