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Math Help - Field and vector space

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Field and vector space

    Let F be a field with  q elements and let V be a vector space above F with dimension of  n .

    1. How many different basis do V have?
    2. How many subspaces of dimension of  k do V have?
    Last edited by Also sprach Zarathustra; December 2nd 2009 at 03:23 PM.
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  2. #2
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let F be a field with  q elements and let V be a vector space above F with dimension of  n .

    1. Who many different basis do V have?
    2. Who many subspaces of dimension of  k do V have?

    It is not "who" but "how"...!נו באמת

    Anyway: how many elements in V can be the first element in the basis? (all but the zero vector): how many elements can be the second element in a basis? (all but the scalar multiples of the first one, and there are q multiples like these), etc.

    Try now to do the second part by yourself.

    Tonio
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Shalom! I am mitbaesh beatzmi...
    Can you please write the whole answer for the question.
    1. q^n-1 ?
    2.?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    p.s:
    What is MHF expert mean?
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    Quote Originally Posted by Also sprach Zarathustra View Post
    p.s:
    What is MHF expert mean?
    Math Help Forum Expert.

    Tonio
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Shalom! I am mitbaesh beatzmi...
    Can you please write the whole answer for the question.
    1. q^n-1 ?
    2.?

    No, not the whole answer but some highlights: there are q^n -1 choices for the first vector (since zero cannot be part of a basis), then there are q^n-q choices for the second vector (all the available vectors minus those that are a scalar multiple of the first one), then there are q^n-q^2 choices for the third one (all the possible vectors minus all the scalar multiples of the first two), etc.
    All in, there are (q^n-1)(q^n-q)(q^n-q^2)...try to end the argument by yourself now.

    Tonio

    Ps ...הרבה יותר כיף לגלות את הפתרון בעצמך, אפילו אם זה קשה, מאשר לקבל את הכל מוכן
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    MHF Contributor Also sprach Zarathustra's Avatar
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    And what about 2?
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    Quote Originally Posted by tonio View Post
    no, not the whole answer but some highlights: There are q^n -1 choices for the first vector (since zero cannot be part of a basis), then there are q^n-q choices for the second vector (all the available vectors minus those that are a scalar multiple of the first one), then there are q^n-q^2 choices for the third one (all the possible vectors minus all the scalar multiples of the first two), etc.
    All in, there are (q^n-1)(q^n-q)(q^n-q^2)...try to end the argument by yourself now.

    Tonio

    ps ...הרבה יותר כיף לגלות את הפתרון בעצמך, אפילו אם זה קשה, מאשר לקבל את הכל מוכן
    גם אתה מישראל?
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  9. #9
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    Quote Originally Posted by Defunkt View Post
    גם אתה מישראל?

    אלא מה? מי עוד כותב עם הסימנים המעצבנים הללו

    Tonio
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  10. #10
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    יפה יפה... אני מניח שכבר סיימת תואר ראשון?
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