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Math Help - vector spaces help

  1. #1
    Junior Member mathmonster's Avatar
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    vector spaces help

    HEY to everyone, never posted here before!!

    if given 2v = v, would it be correct, to prove v = 0, by doing the following, (using the axioms of scalar multiplication and addition for a k-vector space)

    v = v + (v - v) = (v + v) - v = 2v - v, so

    as 2v=v then

    2v-v = v-v = 0.

    thanl u in advance for ur help!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathmonster View Post
    HEY to everyone, never posted here before!!

    if given 2v = v, would it be correct, to prove v = 0, by doing the following, (using the axioms of scalar multiplication and addition for a k-vector space)

    v = v + (v - v) = (v + v) - v = 2v - v, so

    as 2v=v then

    2v-v = v-v = 0.

    thanl u in advance for ur help!
    I would do things slightly differently:
    2v = v

    We know that the additive inverse of v exists and is -v. Thus
    2v + (-v) = v + (-v)

    v = 0

    -Dan
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  3. #3
    Junior Member mathmonster's Avatar
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    thanks, did my way still make sense though?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathmonster View Post
    thanks, did my way still make sense though?
    Yes. It's simply that my method (which uses the same principle that yours did) is a bit shorter is all.

    -Dan
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  5. #5
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    Hello, mathmonster!

    Welcome aboard . . .


    If given 2v = v, would it be correct, to prove v = 0, by doing the following,
    using the axioms of scalar multiplication and addition for a k-vector space?

    . . v \;= \;v + (v - v)

    . . . . = \;(v + v) - v

    . . . . = \;2v - v . . Since 2v = v,
    . . . . . .
    . . . . = \;\;v - v

    . . . . =\quad0

    Sure looks good to me!

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