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The following proof is solely based on vector space axioms. Axiom names are italicised. They are defined in Wikipedia (vector space).

Multiplying by zero vector yields zero vector.

Let $V$ be a vector space over a field $F$. Let $a \in F$. \begin{array}{lrll} \text{By} & \dots & \text{we denote} & \dots \\ \hline & (-a) && \text{an additive inverse of $a$ in $F$.} \\ & 1 && \text{a multiplicative identity element in $F$.} \\ & 0 && \text{an additive identity element in $F$.} \\ & \mathbf{0} && \text{an additive identity element in $V$.} \\ \end{array} We want to prove that $$a \mathbf{0} = \mathbf{0}.$$ Proof. \begin{align*} \mathbf{0} &= 1 \mathbf{0} && \text{by }\textit{Identity element of scalar multiplication} \\ &= (a + (-a)) \mathbf{0} && \text{by }\textit{Inverse elements of field addition} \\ &= a \mathbf{0} + (-a) \mathbf{0} && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\ &= a (\mathbf{0} + \mathbf{0}) + (-a) \mathbf{0} && \text{by }\textit{Identity element of vector addition} \\ &= (a \mathbf{0} + a \mathbf{0}) + (-a) \mathbf{0} && \text{by }\textit{Distrib. of scalar mult. (vector addition)} \\ &= a \mathbf{0} + (a \mathbf{0} + (-a) \mathbf{0}) && \text{by }\textit{Associativity of vector addition} \\ &= a \mathbf{0} + (a + (-a)) \mathbf{0} && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\ &= a \mathbf{0} + 0 \mathbf{0} && \text{by }\textit{Inverse elements of field addition} \\ &= (a + 0) \mathbf{0} && \text{by }\textit{Distrib. of scalar mult. (field addition)} \\ &= a \mathbf{0} && \text{by }\textit{Identity element of field addition} \end{align*} QED

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    $\begingroup$ In the 2nd line, you seem to be using $a+(-a)=1$. $\endgroup$
    – user84413
    Aug 20, 2014 at 17:57
  • $\begingroup$ @user84413 Yes, my mistake! :-( Thanks $\endgroup$
    – DracoMalfoy
    Aug 20, 2014 at 18:03
  • $\begingroup$ ... looks like this can be fixed by replacing 1 with 0 (scalar) in the first line. (math.stackexchange.com/q/893350/138360) $\endgroup$
    – DracoMalfoy
    Aug 20, 2014 at 18:11
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    $\begingroup$ This is all right to do, if you have already shown that $0v=\mathbf{0}$. You might want to use Andre Nicolas' suggestion, though. $\endgroup$
    – user84413
    Aug 20, 2014 at 18:14

1 Answer 1

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Why not $a(\mathbf{0})=a(\mathbf{0}+\mathbf{0})=a(\mathbf{0})+a(\mathbf{0})$; now add the inverse of $a(\mathbf{0})$ to both sides?

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  • $\begingroup$ Actually, I was thinking about that. Out of personal preference - somehow - I wanted to transform a single term (instead of an identity). $\endgroup$
    – DracoMalfoy
    Aug 20, 2014 at 18:23
  • $\begingroup$ Unfortunately that calculation had a glitch early on. I do not see any obvious way of keeping the same basic structure. $\endgroup$
    – André Nicolas
    Aug 20, 2014 at 18:34
  • $\begingroup$ What about the fix suggested above? $\endgroup$
    – DracoMalfoy
    Aug 20, 2014 at 18:39
  • $\begingroup$ Yes, that will work. In a sense it is a longer version of the solution above, since instead of adding the addiive inverse of $a(\mathbf 0)$, you are adding $(-a)(\mathbf{0})$, which is the additive inverse. You are also using a presumably previously proved result $0\mathbf{v}=\mathbf{0}$. $\endgroup$
    – André Nicolas
    Aug 20, 2014 at 18:44

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