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Thread: Help with linear transformation / subspace proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Help with linear transformation / subspace proof

    Q: Let $\displaystyle T:V\rightarrow{W}$ be a linear transformation, and let $\displaystyle U$ be a subspace of $\displaystyle W$. Prove that the set $\displaystyle T^{-1}(U)=\{\vec{v}\in{V} | T(v)\in{U}\}$ is a subspace of $\displaystyle V$. What is $\displaystyle T^{-1}(U)$ if $\displaystyle U=\{\vec{0}\}$ ?

    A: I dont have much, so I will show my train of thought...

    First, I want to show $\displaystyle T^{-1}(U)$ is nonempty. I can do this by simply noting $\displaystyle U$ is a subspace; therefore, $\displaystyle T^{-1}(U)$ is nonempty. I don't know if I am reading the set correctly. Do I wan't to show the range of $\displaystyle T^{-1}(U)$ is nonempty?

    I am confused by what $\displaystyle T^{-1}(U)$ represents.

    Does it mean:

    $\displaystyle T(\vec{v})=c_{1}T(\vec{v_{1}})+c_{2}T(\vec{v_{2}}) +
    ...+c_{k}T(\vec{v_{k}})\in{U}$

    $\displaystyle T^{-1}(T(\vec{v}))=\vec{v}=c_{1}\vec{v_{1}}+c_{2}\vec{ v_{2}}+
    ...+c_{k}\vec{v_{k}}\in{V}$

    ???

    Thanks
    Last edited by Danneedshelp; Jul 27th 2009 at 11:04 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Danneedshelp View Post
    Q: Let $\displaystyle T:V\rightarrow{W}$ be a linear transformation, and let $\displaystyle U$ be a subspace of $\displaystyle W$. Prove that the set $\displaystyle T^{-1}(U)=\{\vec{v}\in{V} | T(v)\in{V}\}$ is a subspace of $\displaystyle V$. What is $\displaystyle T^{-1}(U)$ if $\displaystyle U=\{\vec{0}\}$ ?

    A: I dont have much, so I will show my train of thought...

    First, I want to show $\displaystyle T^{-1}(U)$ is nonempty. I can do this by simply noting $\displaystyle U$ is a subspace; therefore, $\displaystyle T^{-1}(U)$ is nonempty. I don't know if I am reading the set correctly. Do I wan't to show the range of $\displaystyle T^{-1}(U)$ is nonempty?

    I am confused by what $\displaystyle T^{-1}(U)$ represents.

    Does it mean:

    $\displaystyle T(\vec{v})=c_{1}T(\vec{v_{1}})+c_{2}T(\vec{v_{2}}) +
    ...+c_{k}T(\vec{v_{k}})\in{U}$

    $\displaystyle T^{-1}(T(\vec{v}))=\vec{v}=c_{1}\vec{v_{1}}+c_{2}\vec{ v_{2}}+
    ...+c_{k}\vec{v_{k}}\in{V}$

    ???

    Thanks

    $\displaystyle T^{-1}U$ is the inverse image of $\displaystyle U$ under $\displaystyle T$. That is it is the set of all elements $\displaystyle x$ of W such that $\displaystyle Tx\in U$.

    That $\displaystyle T^{-1}U$ is non-empty is shown as $\displaystyle T0_V=0_W $ that is for any linear transformation the zero vector maps to the zero vector. Then since $\displaystyle 0_W\in U$ we have $\displaystyle 0_V\in T^{-1}U$

    To show that $\displaystyle T^{-1}U$ is a subspace of $\displaystyle V$ it is sufficient to show that for any scalars $\displaystyle \alpha_1$, $\displaystyle \alpha_2$ and $\displaystyle x_1, x_2 \in T^{-1}U$ that:

    $\displaystyle
    \alpha_1x_1+\alpha_2x_2 \in T^{-1}U
    $

    CB
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by CaptainBlack View Post

    To show that $\displaystyle T^{-1}U$ is a subspace of $\displaystyle V$ it is sufficient to show that for any scalars $\displaystyle \alpha_1$, $\displaystyle \alpha_2$ and $\displaystyle x_1, x_2 \in T^{-1}U$ that:

    $\displaystyle
    \alpha_1x_1+\alpha_2x_2 \in T^{-1}U
    $

    CB
    Is that all that needs to be shown or do I have to further explain this?

    Thank you
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