# Not a vector space

• Feb 1st 2011, 01:51 AM
I-Think
Not a vector space
Show that if $V$ is a set with 8 elements, then no matter how we de fine 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?
• Feb 1st 2011, 02:25 AM
tonio
Quote:

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
• Feb 1st 2011, 06:16 PM
Drexel28
Quote:

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 $\#(V)=3^{k}$....