**Proof**: *(related to "Simple Consequences from the Definition of a Vector Space")*

Let $V$ be a vector space. Since, by definition, all vectors $x\in V$ form an Abelian group, the statements 1 to 8 follow immediately as a corollary from the group axioms. The statements 9 to 12 remain to be shown.

Ad 9)

- Since $-1=1\cdot (-1)$ in the field $F$, the axioms of scalar multiplication show that $-1x=(1\cdot (-1))x=1\cdot ((-1)x)).$
- Since by these axioms, $1$ is the neutral element of the scalar multiplication in $V$, it follows that $-1x=(-1)x.$

Ad 10)

- Since $0=1+(-1)$ in the field $F$, the axioms of scalar multiplication demonstrate that $0x=(1+(-1))x=x+(-x)=o,$ in which we have used the rules no. 2 and no. 9.

Ad 11)

- By no.2 we have $x+(-x)=o$ for all vectors $x\in V.$
- Thus, by the axioms of scalar multiplication we have $\lambda o=\lambda(x+(-x))=\lambda x + (-\lambda x)=o$ for all $\lambda\in F.$

Ad 12)

No. 10 and/or no. 11 imply $\lambda x=o.$ The converse remains to be shown, so assume $\lambda x=o.$ We want to show that then, at least one of $\lambda$ or $x$ must be zero.

- Assume $\lambda\neq 0.$ Then $\lambda^{-1}\cdot\lambda=1.$ Thus, multiplying both sides of the above equation by $\lambda^{-1}$ and applying the axioms of scalar multiplication as well as no. 11 we get $$\begin{array}{rcl}\lambda^{-1}\cdot(\lambda x)&=&\lambda^{-1} o\\(\lambda^{-1}\cdot\lambda)x&=&o\\1x&=&o\\x&=&o.\end{array}$$
- Now, assume $x\neq o.$ Note that $\lambda x=o=\lambda (x+(-x))=\lambda x+(-\lambda x)=o+(-\lambda x)=-\lambda x.$
- Since $\lambda x=-\lambda x$ and $x\neq o$, we must have $\lambda=-\lambda$, which is only true for $\lambda=0.$

q.e.d

| | | | created: 2018-10-11 23:38:22 | modified: 2018-10-11 23:41:57 | by: *bookofproofs* | references: [577], [7937]

(none)

[7937] **Knabner, P; Barth, W.**: “Lineare Algebra – Grundlagen und Anwendungen”, Springer Spektrum, 2013

[577] **Knauer Ulrich**: “Diskrete Strukturen – kurz gefasst”, Spektrum Akademischer Verlag, 2001

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