Exercise [13.02]

Sameed Zahoor · 24 Aug 2008, 10:11

Let V be a vector space.Then,
1.V is closed under vector addition.(The sum of two vectors is again a vector.)
2.For any two vectors $\xi$ and $\eta$ in V we have by definition,
$\xi +\eta=\eta +\xi$ (Commutativity)
3.For any three $\xi$ , $\eta$ and $\zeta$ in V,we again have by definition
$\xi +(\eta +\zeta)=(\xi +\eta) +\zeta$ (Associativity)
4.There exists a null vector such that,
$0+\xi=\xi + 0=\xi$ for all $\xi$ in V.(Existence of identity)
5.There exists a 'negative' vector $-\xi$ for any $\xi$ in V such that,
$-\xi +\xi=\xi +(-\xi)=0$ (Existence of inverse)
Thus,V satisfies all the requirements of an abelian group.

Exercise [13.02]

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