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Math Help - Llinear transformations: Help with proving linear independency

  1. #1
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    Llinear transformations: Help with proving linear independency

    Hi guys.

    I had the following statement to prove:

    Let \varphi :V\rightarrow W be a linear transformation, and v_1, v_2,...v_n\in V a set of vectors.
    If v_1, v_2,...v_n are linear independent, and \varphi is one-to-one, then \varphi(v_1), \varphi(v_2),...,\varphi(v_n) are also linear independent.

    that's how I proved it:
    v_1, v_2,...v_n are linear independent, so if:

    \alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n=0

    then: \alpha_1=\alpha_2=...=\alpha_n=0

    now, since \varphi is a linear transformation, it holds that \varphi(\vec{0}_V)=\vec{0}_W, therefore:

    \varphi(\vec{0}_V)=\varphi(\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n)=
    \alpha_1\varphi(v_1)+\alpha_2\varphi(v_2)+...+ \alpha_n\varphi(v_n)=\vec{0}_W

    and since we know that \alpha_1=\alpha_2=...=\alpha_n=0, then \varphi(v_1), \varphi(v_2),...,\varphi(v_n) are linear independent.

    my question here is: why should I care that \varphi is one-to-one?

    any help would be greatly appreciated.
    Last edited by Stormey; March 21st 2013 at 10:09 AM.
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  2. #2
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    Re: Llinear transformations: Help with proving linear independency

    Quote Originally Posted by Stormey View Post
    Let \varphi :V\rightarrow W be a linear transformation, and v_1, v_2,...v_n\in V a set of vectors.
    If v_1, v_2,...v_n are linear independent, and \varphi is one-to-one, then \varphi(v_1), \varphi(v_2),...,\varphi(v_n) are also linear independent.

    now, since \varphi is a linear transformation, it holds that \varphi(\vec{0}_V)=\vec{0}_W, therefore:

    \varphi(\vec{0}_V)=\varphi(\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n)=
    \alpha_1\varphi(v_1)+\alpha_2\varphi(v_2)+...+ \alpha_n\varphi(v_n)=\vec{0}_W

    and since we know that \alpha_1=\alpha_2=...=\alpha_n=0, then \varphi(v_1), \varphi(v_2),...,\varphi(v_n) are linear independent.

    my question here is: why should I care that \varphi is one-to-one?

    any help would be greatly appreciated.
    What you did is not a proof.
    You should start with, suppose that \alpha_1\varphi(v_1)+\alpha_2\varphi(v_2)+...+ \alpha_n\varphi(v_n)=\vec{0}_W
    From which you must show that each \alpha_k must equal zero.

    Now as you said \varphi(\vec{0}_V)=\varphi(\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n) is also that sum.

    But that means \vec{0}_V=\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n because \varphi is one-to-one.

    Are you done? Why?
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    Re: Llinear transformations: Help with proving linear independency

    Hi Plato, and thanks for the help.

    Yeah, we're done, if \vec{0}_V=\alpha_1 v_1+\alpha_2 v_2+...+\alpha_nv_n, then \alpha_1=\alpha_2=...=\alpha_n=0
    because v_1, v_2,...v_n are linear independent, and therefore \varphi(v_1), \varphi(v_2),...,\varphi(v_n) are linear independent.

    what was wrong with mine?
    why doesn't it considered a valid proof?
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    Re: Llinear transformations: Help with proving linear independency

    Quote Originally Posted by Stormey View Post
    what was wrong with mine?
    why doesn't it considered a valid proof?
    You started out with what you wanted to prove. So you went nowhere.
    That is why you never had to use one-to-one.
    You are to show that \left\{ {\varphi ({v_k})} \right\}_{k = 1}^n is an independent set.
    So you start with a linear combination equaling the zero vector. Not the other way round.
    Thanks from Stormey
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