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

Vector spaces - Multiplying by zero scalar yields zero vector

\begin{array}{lrll} \text{Let} & \dots & \text{be} & \dots \\ \hline & F && \text{a field.} \\ & V && \text{a vector space over $F$.} \\ & 0 && \text{an identity element of addition of $F$.} \\ & \mathbf{0} && \text{an identity element of addition of $V$.} \\ & \mathbf{v} && \text{an arbitrary vector in $V$.} \\ \end{array}

$$\text{Then, }0\mathbf{v} = \mathbf{0}.$$

Proof. We will denote by $1$ an identity element of scalar multiplication; we will denote by $(-\mathbf{v})$ an additive inverse of $\mathbf{v}$. \begin{align*} 0\mathbf{v} &= 0\mathbf{v} + \mathbf{0} && \text{by }\textit{Identity element of vector addition} \\ &= 0\mathbf{v} + (\mathbf{v} + (-\mathbf{v})) && \text{by }\textit{Inverse elements of vector addition} \\ &= (0\mathbf{v} + \mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Associativity of vector addition} \\ &= (0\mathbf{v} + 1\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Identity element of scalar multiplication} \\ &= ((0 + 1)\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Distributivity of scalar multiplication (field addition)} \\ &= ((1 + 0)\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Commutativity of field addition} \\ &= (1\mathbf{v}) + (-\mathbf{v}) && \text{by }\textit{Identity element of field addition} \\ &= \mathbf{v} + (-\mathbf{v}) && \text{by }\textit{Identity element of scalar multiplication} \\ &= \mathbf{0} && \text{by }\textit{Inverse elements of vector addition} \\ \end{align*} QED

  • 3
    $\begingroup$ Looks great to me. $\endgroup$ – Paul Sundheim Aug 10 '14 at 20:30
  • $\begingroup$ How do you like the "Let...be..."-table? $\endgroup$ – DracoMalfoy Aug 10 '14 at 20:35
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    $\begingroup$ For speed, and I think greater naturalness, try $0v=(0+0)v=0v+0v$. Now add the additive inverse of $0v$ to both sides. $\endgroup$ – André Nicolas Aug 10 '14 at 20:38
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    $\begingroup$ @draco malfoy Usually the theorem you want to prove contains, at least implicitly, most of the items in the table and it isn't stated in that way. So, rather than using a table, it usually suffices to just start your proof with "Let $v$ be an arbitrary element of the vector space $V$. Also, at this level, the equations without the explanations is enough. But you would need to check with your professor about that. The theorem might read as "Let V be a vector space over a field F. For any vector v in V, and then your statement that follows Then. $\endgroup$ – Paul Sundheim Aug 10 '14 at 21:10
  • $\begingroup$ As a follow-up to Andre's comment, you could shorten your proof (but retain the style you have) by replacing $(v+(-v))$ by $(0v+(-0v))$ in line 2. $\endgroup$ – user84413 Aug 10 '14 at 22:26

To shorten the proof, we may write as suggested by André Nicolas,

Proof. Let $(V,+,\cdot)_F$ be a vector space over the field $F$. We wish to show that $\forall v\in V$ one has $0\cdot v=\mathbf{0}$, where $0$ is the zero scalar and $\mathbf{0}$ is the zero vector. Let $v$ be an element of the vector space $V$;

By one of the axioms of field addition, $$0\cdot v=(0+0)\cdot v.$$ Since scalar multiplication is distributive over addition, $$(0+0)\cdot v=0\cdot v+0\cdot v.$$ From the previous two equalities we conclude that $$0\cdot v=0\cdot v+0\cdot v.$$ Adding to both sides the inverse element for addition of $0\cdot v$, which we'll denote by $-0\cdot v$: $$0\cdot v+(-0\cdot v)=0\cdot v+0\cdot v+(-0\cdot v).$$ By the inverse axiom, $$\mathbf{0}=0\cdot v+\mathbf{0},$$ hence by the identity axiom, $$0\cdot v=\mathbf{0}.\tag*{$\square$}$$

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