Not a vector space

**I-Think** · February 1st 2011, 01:51 AM

Show that if $V$ is a set with 8 elements, then no matter how we define addition on $V$
and multiplication by scalars from $Z_3$ , $V$ cannot be made a vector space over $Z_3$ .

I'm thinking that I demonstrate that one of the properties of vector spaces, (closure, commutative etc) doesn't hold.
How do I show this for all different ways of defining addition and scalar multiplication though?

**tonio** · February 1st 2011, 02:25 AM

Originally Posted by I-Think

Show that if $V$ is a set with 8 elements, then no matter how we define addition on $V$
and multiplication by scalars from $Z_3$ , $V$ cannot be made a vector space over $Z_3$ .

I'm thinking that I demonstrate that one of the properties of vector spaces, (closure, commutative etc) doesn't hold.
How do I show this for all different ways of defining addition and scalar multiplication though?

Hint: for any $v\in V$ , it'd be true that $0=3\cdot v = v+v+v$

Tonio

**Drexel28** · February 1st 2011, 06:16 PM

Originally Posted by I-Think

Show that if $V$ is a set with 8 elements, then no matter how we define addition on $V$
and multiplication by scalars from $Z_3$ , $V$ cannot be made a vector space over $Z_3$ .

I'm thinking that I demonstrate that one of the properties of vector spaces, (closure, commutative etc) doesn't hold.
How do I show this for all different ways of defining addition and scalar multiplication though?

Let $\dim V=k$ then $V\cong \mathbb{Z}_3^k$ and thus $\#(V)=3^{k}$ ....

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