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Math Help - Question on subspace

  1. #1
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    Question on subspace

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  2. #2
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    Re: Question on subspace

    I assume the question deals with vector spaces over real numbers. Is W closed under multiplication by a real number?
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    Re: Question on subspace

    Yes, it is deal with vector spaces over real number. For a vector to be subspaces, it must be closed under addition and scalar multiplication.

    (a) (a1, b1) + (a2, b2) = (a1 + a2, b1 + b2) => closed under addition
    k(a, b) = (ka, kb) => closed under scalar multiplication

    But i do think that if a>= 0, then k(a, b) = (ka, kb) is not closed under scalar multiplication when k is negative value so it is not subspace of V.


    (b) (a1, c1, c1) + (a2, c2, c2) = (a1 + a2, c1 + c2, c1 + c2) => closed under addition
    k(a, c, c) = (ka, kc, kc) => closed under scalar multiplication

    Similarly, a^2 + b^2 + c^2 <= 1 could mean a, b, c might be 0, 0<= a <= 1 or -1 <= a <= 0. But i do not know why b is there since this question does not have value b.
    and if a is 1, then a1 + a2 = 2 which is not closed under addition so it is not subspace too.

    I hope that my working is correct. But i doubt that i explain it in correct way. Or maybe i shall prove that with contradiction.
    (1, 0, 0) + (1, 0, 0) = (2, 0, 0) which is not closed under addition.

    Could you further elaborate on that?
    Last edited by whyyie; June 6th 2012 at 03:28 AM.
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  4. #4
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    Re: Question on subspace

    Quote Originally Posted by whyyie View Post
    But i do think that if a>= 0, then k(a, b) = (ka, kb) is not closed under scalar multiplication when k is negative value so it is not subspace of V.
    Yes. I would note that only a set can be closed under an operation. It is not correct to say that k(a, b) = (ka, kb), which is an equality and not a set, is closed under scalar multiplication.

    Quote Originally Posted by whyyie View Post
    Similarly, a^2 + b^2 + c^2 <= 1 could mean a, b, c might be 0, 1 or -1. But i do not know why b is there since this question does not have value b.
    I am not sure what you mean. Why do you set apart b? Does the question have the value of a or c?

    Quote Originally Posted by whyyie View Post
    and if a is 1, then a1 + a2 = 2 which is not closed under addition
    What are a1 and a2? Why is a1 + a2 = 2? Again, "... = ..., which is not closed" is incorrect.

    The claim "W is closed under scalar multiplication" means "For all real numbers k and all vectors (a, b, c) ∈ W, (ka, kb, kc) ∈ W." To form its negation, switch the quantifiers and add the negation to the final part: "There exists a real number k and a vector (a, b, c) ∈ W such that (ka, kb, kc) ∉ W." Can you find specific k, a, b, c for which this negation is true?
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    Re: Question on subspace

    it appears there is a typo in the second part of the problem, and i believe it should be triples of the form (a,b,c) instead of (a,c,c).

    i also suggest using a k much larger than 1 as a scalar for a counter-example for the second part.
    Thanks from emakarov and HallsofIvy
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  6. #6
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    Re: Question on subspace

    Quote Originally Posted by Deveno View Post
    it appears there is a typo in the second part of the problem, and i believe it should be triples of the form (a,b,c) instead of (a,c,c).
    Thanks for noticing.

    Quote Originally Posted by Deveno View Post
    i also suggest using a k much larger than 1 as a scalar for a counter-example for the second part.
    This reminds me what my calculus teacher used to say: "Let ε be arbitrarily small: for example, 100."
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  7. #7
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    Re: Question on subspace

    one of my favorite jokes is:

    1+1 = 3, for large values of 1.
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