# Vector spaces - Multiplying by $-1$ yields inverse element of vector addition.

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The following proof is based on vector space related axioms. Axiom names are italicised. They are defined in Wikipedia (see vector space article). Additionally, this trivial result is used.

## Multiplying by $-1$ yields inverse element of vector addition.

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

• Yeah, that's fine – Hagen von Eitzen Aug 20 '14 at 16:18
• Your proof is good. I will just comment that in the right-hand column in your table, all four "a"/"an" should be "the", because the three identities and -1 are all unique. – Bungo Aug 20 '14 at 16:25
• @Bungo I did not use "the" because uniqueness is not explicit in the axioms (in Wikipedia). – DracoMalfoy Aug 20 '14 at 16:28
• @draco - OK, fair enough :-) – Bungo Aug 20 '14 at 17:06
• i.e. scale $\,\bf v\,$ by $\ 0\, =\, -1 + 1\ \$ – Bill Dubuque Aug 20 '14 at 17:50