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Thread: prove T(V) is a subspace

  1. #1
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    prove T(V) is a subspace

    Let $\displaystyle T : R^n --> R^m$be a linear transformation, if V is a subspace of $\displaystyle R^n$ and
    T(V) = {T(v): $\displaystyle v \in R^n$},
    prove that T(V) is a subspace of $\displaystyle R^m$.
    Sry, I know this is an easy question, but i am kindof confused with the definition of subspace
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  2. #2
    Member Black's Avatar
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    Refer to this thread.
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  3. #3
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    so is this right

    If$\displaystyle v, w \in V$, then T(v) = v and T(w) = w so we get
    T(v + w) = T(v) + T(w) = v + w
    which means that $\displaystyle v+w \in V$ and condition (1) holds.
    For any scalar c,
    T(cv) = cT(v) = cv
    so that $\displaystyle cv \in V$and condition (2) holds so V is a subspace

    is $\displaystyle R^m$same as $\displaystyle R^n$?
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  4. #4
    Member Black's Avatar
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    Quote Originally Posted by wopashui View Post
    so is this right

    If$\displaystyle v, w \in V$, then T(v) = v and T(w) = w so we get
    T(v + w) = T(v) + T(w) = v + w
    which means that $\displaystyle v+w \in V$ and condition (1) holds.
    For any scalar c,
    T(cv) = cT(v) = cv
    so that $\displaystyle cv \in V$and condition (2) holds so V is a subspace

    is $\displaystyle R^m$same as $\displaystyle R^n$?
    Not in general. If $\displaystyle n=m$, then $\displaystyle \mathbb{R}^n=\mathbb{R}^m$ ; they are not the same otherwise.

    To show that $\displaystyle T(V)$ is a subspace of $\displaystyle \mathbb{R}^m$, you have to prove three conditions:

    (1) $\displaystyle 0_{\mathbb{R}^m} \in T(V),$
    (2) If $\displaystyle u,v \in T(V)$, then $\displaystyle u+v \in T(V),$
    (3) If $\displaystyle \lambda \in \mathbb{R}$ and $\displaystyle u \in T(V)$, then $\displaystyle \lambda u \in T(V)$.

    For (1), $\displaystyle 0_{\mathbb{R}^m}=T\left(0_{\mathbb{R}^n}\right) \in T(V)$.

    For (2), if $\displaystyle u,v \in T(V)$, then pick $\displaystyle u_0,v_0 \in \mathbb{R}^n$ such that $\displaystyle T(u_0)=u$ and $\displaystyle T(v_0)=v$. Then

    $\displaystyle u+v=T(u_0)+T(v_0)=T(u_0+v_0) \in T(V).$

    And for (3), you just use preservation of scalar multiplication.
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  5. #5
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    Quote Originally Posted by Black View Post
    Not in general. If $\displaystyle n=m$, then $\displaystyle \mathbb{R}^n=\mathbb{R}^m$ ; they are not the same otherwise.

    To show that $\displaystyle T(V)$ is a subspace of $\displaystyle \mathbb{R}^m$, you have to prove three conditions:

    (1) $\displaystyle 0_{\mathbb{R}^m} \in T(V),$
    (2) If $\displaystyle u,v \in T(V)$, then $\displaystyle u+v \in T(V),$
    (3) If $\displaystyle \lambda \in \mathbb{R}$ and $\displaystyle u \in T(V)$, then $\displaystyle \lambda u \in T(V)$.

    For (1), $\displaystyle 0_{\mathbb{R}^m}=T\left(0_{\mathbb{R}^n}\right) \in T(V)$.

    For (2), if $\displaystyle u,v \in T(V)$, then pick $\displaystyle u_0,v_0 \in \mathbb{R}^n$ such that $\displaystyle T(u_0)=u$ and $\displaystyle T(v_0)=v$. Then

    $\displaystyle u+v=T(u_0)+T(v_0)=T(u_0+v_0) \in T(V).$

    And for (3), you just use preservation of scalar multiplication.
    so for (3) if $\displaystyle u\in T(v)$, $\displaystyle c \in R$, then we have
    cu = $\displaystyle cT(u') \in T(v) $

    right?
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  6. #6
    Member Black's Avatar
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    Quote Originally Posted by wopashui View Post
    so for (3) if $\displaystyle u\in T(v)$, $\displaystyle c \in R$, then we have
    cu = $\displaystyle cT(u')$ = T(cu') $\displaystyle \in T(v)$

    right?
    Need to add that part.
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  7. #7
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    Quote Originally Posted by Black View Post
    Need to add that part.
    You certainly provided a very good description. Thanks a lot.
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