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Thread: First Isomorphism Theorem

  1. #1
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    First Isomorphism Theorem

    Let S,T be subspaces of the vector spaces V ,U (respectively), and let

    $\displaystyle T : V U -> (V/S) (U/T)$ be the linear map defined by $\displaystyle T(x, y) = (x +S, y + T).$ Find the kernel of T and prove that

    $\displaystyle (V U)/(S T) (~ =) (V/S) (W/T).$

    Plx help...
    Last edited by mathbeginner; Nov 3rd 2010 at 06:58 PM. Reason: edit
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  2. #2
    Senior Member roninpro's Avatar
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    First, do you mean to define $\displaystyle T(x, y) = (x +S, y + T)$?

    I think that your first step is to determine the kernel of $\displaystyle T$. If $\displaystyle T(x,y)=(0 +S, 0 +T)$, what can be said about $\displaystyle x$ and $\displaystyle y$?
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  3. #3
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    Quote Originally Posted by roninpro View Post
    First, do you mean to define $\displaystyle T(x, y) = (x +S, y + T)$?

    I think that your first step is to determine the kernel of $\displaystyle T$. If $\displaystyle T(x,y)=(0 +S, 0 +T)$, what can be said about $\displaystyle x$ and $\displaystyle y$?
    is it Ker T= T(0,0)=(0+S, 0+T) =(S,T)
    therefor Ker T = (S,T)?

    but I still don't know how to prove the following....
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathbeginner View Post
    is it Ker T= T(0,0)=(0+S, 0+T) =(S,T)
    therefor Ker T = (S,T)?

    but I still don't know how to prove the following....
    You have that $\displaystyle F:V\boxplus U\to (V/S)\boxplus (U/T)x,y)\mapsto \left(x+S,y+T)$ (I use $\displaystyle \boxplus$ to differentiate between external and internal direct sum), this is evidently a surjective linear homomorphism. Then, we are trying to look for what values $\displaystyle (x,y)$ is $\displaystyle T(x,y)=...$...equals what? We're trying to look for the origin in the vector spaces $\displaystyle V/S$ and $\displaystyle U/T$ but these are precisely $\displaystyle S$ and $\displaystyle T$ respectively. So, for what $\displaystyle x$ is $\displaystyle x+S=S$ and for what $\displaystyle y$ is $\displaystyle y+T=T$? It is fairly clear that these statements are true if and only if $\displaystyle x\in S$and $\displaystyle y\in T$. Thus, a quick proof shows that $\displaystyle \ker F=S\boxplus T$ and so by the F.I.T. $\displaystyle \left(V\boxplus U\right)/\left(S\boxplus T\right)\cong \left(V/S\right)\boxplus\left(U/T\right)$
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    I don't reli get why Ker F = S X T and how to prove that???
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mathbeginner View Post
    I don't reli get why Ker F = S X T and how to prove that???
    Come on friend. Review your definitions. You're trying to show that $\displaystyle \ker F=S\boxplus T$ which is equivalent to showing $\displaystyle F(x,y)=(S,T)\Leftrightarrow (x,y)\in S\boxplus T$. I give the solution below "hidden". Try your damnedest, and only look at it as a last result.
    [spoil]Clearly if $\displaystyle (x,y)\in S\boxplus T$ then $\displaystyle F(x,y)=(x+S,y+T)=(S,T)$ since $\displaystyle x+S=\left\{x+s:s\in S\right\}=S$ since $\displaystyle S\leqslant V$ and similarly $\displaystyle y+T=T$. Conversely, suppose that $latex F(x,y)=(S,T)[/tex] then $\displaystyle x+S=S$. But, note that $\displaystyle x\in x+S$ since $\displaystyle 0\in S$ and so $\displaystyle x\in S$. Similarly $\displaystyle y+T=T\implies y\in T$ and so $\displaystyle F(x,y)=(S,T)\implies (x,y)\in (S,T)$. Combining these we get $\displaystyle F(x,y)=(S,T)\Leftrightarrow (x,y)\in S\boxplus T$[/spoil]

    EDIT: I can't figure out how to hide it...so work on the honor code haha
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    EDIT: I can't figure out how to hide it...so work on the honor code haha
    [Spoiler] [/Spoiler] You have been gone for so long that you forgot the tags!
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    [Spoiler] [/Spoiler] You have been gone for so long that you forgot the tags!
    Haha, I guess so!
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