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Math Help - subspace and isomorphism

  1. #1
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    subspace and isomorphism

    Let U be a vector space, U x U the product vector space, and

    # = { (u,u)|u in U} ⊆ U x U

    Prove that # is a subspace of U x U and that (U x U) / # (isomorphism) U.
    plx help....
    Last edited by mathbeginner; November 3rd 2010 at 07:55 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathbeginner View Post
    Let U be a vector space, U x U the product vector space, and

    # = { (u,u)|u in U} ⊆ U U.

    Prove that # is a subspace of U x U and that (U x U) / # (isomorphism) U.

    plx help....
    What about the obvious choice of u_1,u_2)\to u_1-u_2" alt="T:U\boxplus U\to Uu_1,u_2)\to u_1-u_2" />. This is evidently surjective since T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U} and it's a homomorphism since T(\alpha(u_1,u_2)+\beta(u_3,u_4))=\left(\alpha u_1+\beta u_3,\alpha u_2+\beta u_4)=(\alpha u_1+\beta u_3)-(\alpha u_2+\beta u_4)=\alpha(u_1-u_2)+\beta(u_3-u_4)=\alpha T((u_1,u_2))+\beta T((u)3,u_4)). I claim that we are now finished. Why?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    What about the obvious choice of u_1,u_2)\to u_1-u_2" alt="T:U\boxplus U\to Uu_1,u_2)\to u_1-u_2" />. This is evidently surjective since T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U} and it's a homomorphism since T(\alpha(u_1,u_2)+\beta(u_3,u_4))=\left(\alpha u_1+\beta u_3,\alpha u_2+\beta u_4)=(\alpha u_1+\beta u_3)-(\alpha u_2+\beta u_4)=\alpha(u_1-u_2)+\beta(u_3-u_4)=\alpha T((u_1,u_2))+\beta T((u)3,u_4)). I claim that we are now finished. Why?
    why we only prove that # is a subspace of U then it is homomorphism
    and I don't reli get why This is evidently surjective since T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U}
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathbeginner View Post
    why we only prove that # is a subspace of U then it is homomorphism
    and I don't reli get why This is evidently surjective since T_{\mid U\boxplus\{\bold{0}\}}=\text{id}_{U}
    Let u\in U then [math}T(u,0)=u-0=u[/tex]
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