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Thread: Not a vector space

  1. #1
    Senior Member I-Think's Avatar
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    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?
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  2. #2
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    Quote Originally Posted by I-Think View Post
    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
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by I-Think View Post
    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 $\displaystyle \#(V)=3^{k}$....
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