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Math Help - dimensional vector space

  1. #1
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    dimensional vector space

    Suppose that K is a finite field with q elements. Let V be an n-dimensional vector space over k. How munch element does v have?
    I guess v will have n-q element, but I am not sure.
    and if I am right how can I show that v have n-q element???

    Pls help!
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  2. #2
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    Quote Originally Posted by mathbeginner View Post
    I guess v will have n-q element, but I am not sure.
    and if I am right how can I show that v have n-q element???

    Pls help!
    How about this is V is 1 dim vector space how many distinct elements are there?

    Okay Now what is V is a 2 dim vector space?

    V is 3 dim?

    .
    .
    .
    n dim

    just follow the pattern
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  3. #3
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    if only for v is n dim then v will have n element,
    but now V is an n-dimensional vector space over k.
    will it change the element in V???
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    Quote Originally Posted by mathbeginner View Post
    if only for v is n dim then v will have n element,
    but now V is an n-dimensional vector space over k.
    will it change the element in V???
    You know that since \dim_{\mathbb{K}} V=n that V\cong \mathbb{K}^n. Now, tell me. In general if a set X has m elements, how many elements does X^m have?
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  5. #5
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    it will also have m element, right?
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  6. #6
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    Quote Originally Posted by mathbeginner View Post
    it will also have m element, right?
    no..now many ordered pairs oj the set {1,2,3} are there?
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    Quote Originally Posted by Drexel28 View Post
    no..now many ordered pairs oj the set {1,2,3} are there?
    3 there are 3 element in the set
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  8. #8
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    Quote Originally Posted by mathbeginner View Post
    3 there are 3 element in the set
    No. There are 9, (i,j),\text{ }i,j=1,2,3. In general if |X|<\infty (there's no need to consider infinite sets for this example) it's true that \left|X^m\right|=|X|^m.
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  9. #9
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    Quote Originally Posted by Drexel28 View Post
    No. There are 9, (i,j),\text{ }i,j=1,2,3. In general if |X|<\infty (there's no need to consider infinite sets for this example) it's true that \left|X^m\right|=|X|^m.
    In general if a set X has m elements, how many elements does X^m have?[/QUOTE]
    so you are saying the ans of that is X^m then is m^m since x have m elements?
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  10. #10
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    Quote Originally Posted by mathbeginner View Post
    so you are saying the ans of that is X^m then is m^m since x have m elements?
    What I'm saying is that if X has n elements then X^m has n^m elements.
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  11. #11
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    Quote Originally Posted by Drexel28 View Post
    You know that since \dim_{\mathbb{K}} V=n that V\cong \mathbb{K}^n. Now, tell me. In general if a set X has m elements, how many elements does X^m have?
    I also don't really get why \dim_{\mathbb{K}} V=n will become V\cong \mathbb{K}^n
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  12. #12
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    Then look at simple cases. Way back in the first response TheEmptySet asked you "if V is 1 dim vector space how many distinct elements are there?" You did not answer that- instead you responded about n- dimensions.

    If V is a one dimensional vector space, then it has a single basis vector, v, say. All vectors are of the form av where a is from K. If, for example, K is a finite field with 3 elements, 0, 1, 2, say, then the only vectors in the space are 0v, 1v, and 2v- exactly 3 vectors.

    If V is a two dimensional vector space, then a basis contains two vectors, u and v, say. All vectors are of the form au+ bv where a and b are from K. Again, if K has only 0, 1, and 2 then the only vectors are 0u+ 0v, 0u+ 1v, 0u+ 2v, 1u+ 0v, 1u+ 1v, 1u+ 2v, 2u+ 0v, 2u+ 1v, and 2u+ 2v- exactly 9 vectors.

    Do you get the point? How many vectors would there be if V were three dimensional?
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  13. #13
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    Quote Originally Posted by HallsofIvy View Post
    Then look at simple cases. Way back in the first response TheEmptySet asked you "if V is 1 dim vector space how many distinct elements are there?" You did not answer that- instead you responded about n- dimensions.

    If V is a one dimensional vector space, then it has a single basis vector, v, say. All vectors are of the form av where a is from K. If, for example, K is a finite field with 3 elements, 0, 1, 2, say, then the only vectors in the space are 0v, 1v, and 2v- exactly 3 vectors.

    If V is a two dimensional vector space, then a basis contains two vectors, u and v, say. All vectors are of the form au+ bv where a and b are from K. Again, if K has only 0, 1, and 2 then the only vectors are 0u+ 0v, 0u+ 1v, 0u+ 2v, 1u+ 0v, 1u+ 1v, 1u+ 2v, 2u+ 0v, 2u+ 1v, and 2u+ 2v- exactly 9 vectors.

    Do you get the point? How many vectors would there be if V were three dimensional?
    Is it the basis contain 3 vectors and will have 3(vector form K)^3(dim of vector space) am I right?
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  14. #14
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    Quote Originally Posted by mathbeginner View Post
    Is it the basis contain 3 vectors and will have 3(vector form K)^3(dim of vector space) am I right?
    Is this 3\left(\text{number of things in }\mathbb{K}\right)^3 or just \left(\text{number of things in }\mathbb{K}\right)^3?
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  15. #15
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    Yes. And if K contains 3 members, then there would be 3^n members of an n dimensional space.

    If K contains k members, an n dimensional space over K would contain k^n members.
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