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Please rate and comment. I want to improve; constructive criticism is highly appreciated. Please take style into account as well.

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 '14 at 17:57
  • $\begingroup$ @user84413 Yes, my mistake! :-( Thanks $\endgroup$ – DracoMalfoy Aug 20 '14 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 '14 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 '14 at 18:14
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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 '14 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 '14 at 18:34
  • $\begingroup$ What about the fix suggested above? $\endgroup$ – DracoMalfoy Aug 20 '14 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 '14 at 18:44

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