# Not a vector space

• Feb 1st 2011, 01:51 AM
I-Think
Not a vector space
Show that if $\displaystyle V$ is a set with 8 elements, then no matter how we de fine addition on $\displaystyle V$
and multiplication by scalars from $\displaystyle Z_3$, $\displaystyle V$ cannot be made a vector space over $\displaystyle Z_3$.

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

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 $\displaystyle v\in V$, it'd be true that $\displaystyle 0=3\cdot v = v+v+v$

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

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

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