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Math Help - I thought this is done to show that set is a subspace, not a vector space?

  1. #1
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    I thought this is done to show that set is a subspace, not a vector space?

    Question: show that the set L={(x,y,z,w) is in R^4 | x+y-z+w=0} with operations from R^4 is a vector space.

    Answer: show that it's closed under vector addition (between itself and an arbatrary vecotr) and closed under scalar multiplication (between itself and an arbitrary constant)


    Confusion: I thought that's what's done to show something is a subspace? To show something is a vector space, the set must satisfy 10 properties including things like

    \mathbf{u} \oplus \mathbf{v} = \mathbf{v} \oplus \mathbf{u} for u and v in the vector space
    and
    (c+d) \odot \mathbf{u} = c \odot \mathbf{u} \oplus d \odot \mathbf{u}

    so why is it in the answer they only showed that it was closed under scalar multiplication and vector addition and didn't need to show the other properties, for example like the two I list above?
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  2. #2
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    Quote Originally Posted by superdude View Post
    Question: show that the set L={(x,y,z,w) is in R^4 | x+y-z+w=0} with operations from R^4 is a vector space.



    Answer: show that it's closed under vector addition (between itself and an arbatrary vecotr) and closed under scalar multiplication (between itself and an arbitrary constant)

    Confusion: I thought that's what's done to show something is a subspace? To show something is a vector space, the set must satisfy 10 properties including things like


    \mathbf{u} \oplus \mathbf{v} = \mathbf{v} \oplus \mathbf{u} for u and v in the vector space
    and
    (c+d) \odot \mathbf{u} = c \odot \mathbf{u} \oplus d \odot \mathbf{u}

    so why is it in the answer they only showed that it was closed under scalar multiplication and vector addition and didn't need to show the other properties, for example like the two I list above?

    Because:

    (1) A vector subsapce is a vector space in its own right, it is just taken out from a bigger vector space that contains it;

    (2) In this case, the set L is clearly a subset of \mathbb{R}^4 so, unless otherwise given, we can assume the same operations as in this vector space apply and thus to show L is a space is the same as showing it is a subspace of \mathbb{R}^4.

    Tonio
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