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Math Help - Clarification on a problem

  1. #1
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    Clarification on a problem

    Hey guys,
    I have this problem: Prove that U1 := {(x1,...,xn) ∈ F^n | x1 +...+xn = 0} is a subspace of F^n, but U2 := {(x1, . . . , xn) ∈ F^n | x1 + . . . + xn = 1} is not.

    My question is do I go about showing U1 as a subspace and U2 not as a subspace as one would do normally? Or are there special conditions we have to follow because we are dealing with F^n (n-tuples). I tried looking in my book and my book doesn't really show anything on it and I tried google but i can't seem to find what i'm looking for.
    Thank you.
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  2. #2
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    Quote Originally Posted by alice8675309 View Post
    Hey guys,
    I have this problem: Prove that U1 := {(x1,...,xn) ∈ F^n | x1 +...+xn = 0} is a subspace of F^n, but U2 := {(x1, . . . , xn) ∈ F^n | x1 + . . . + xn = 1} is not.

    My question is do I go about showing U1 as a subspace and U2 not as a subspace as one would do normally? Or are there special conditions we have to follow because we are dealing with F^n (n-tuples). I tried looking in my book and my book doesn't really show anything on it and I tried google but i can't seem to find what i'm looking for.
    Thank you.

    Show that U_1 isn't empty and is closed under vector addition and multiplication by scalars, and then show

    either that U_2 isn't closed under neither of both operations above (pretty easy to do), or else show it doesn't

    contain the zero vector (why is this a necessary, though not sufficient, condition for something to be a subspace?)

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    Show that U_1 isn't empty and is closed under vector addition and multiplication by scalars, and then show

    either that U_2 isn't closed under neither of both operations above (pretty easy to do), or else show it doesn't

    contain the zero vector (why is this a necessary, though not sufficient, condition for something to be a subspace?)

    Tonio
    ok got it. Thanks! I was just making sure that it was the same usual steps and not something different because we were dealing with n-tuples. Couldn't remember. Thanks again
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