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Math Help - Kernal and Image of a Linear Transformation

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    Kernal and Image of a Linear Transformation

    Given \{v_1, ... ,v_n\} in a vector space V, define T: \mathbb{R}^n \rightarrow V by T(r_1, ..., r_n) = r_1v_1+...+r_nv_n

    Show that T is one-to-one if and only if \{v_1,...,v_n\} is independent.
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    Quote Originally Posted by Scopur View Post
    Given \{v_1, ... ,v_n\} in a vector space V, define T: \mathbb{R}^n \rightarrow V by T(r_1, ..., r_n) = r_1v_1+...+r_nv_n

    Show that T is one-to-one if and only if \{v_1,...,v_n\} is independent.
    T is a linear transformation. Therefore, T is one-to-one if and only if \ker (T) = \{ \bold{0} \} if and only if r_1\bold{v}_1+...+r_n\bold{v_n} = \bold{0} \implies r_1=r_2=...=r_n=0
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    uhh.. 2nd part i get one and i have the answer to this but.. I don't quite understand onto as well as the kernel.

    2. T is onto iff V = span\{v_1,...,v_n\}
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    Quote Originally Posted by Scopur View Post
    uhh.. 2nd part i get one and i have the answer to this but.. I don't quite understand onto as well as the kernel.

    2. T is onto iff V = span\{v_1,...,v_n\}
    A linear transformation T: \mathbb{R}^n \to V is onto iff for any \bold{v}\in V we can find (r_1,...,r_n)\in \mathbb{R}^n so that T(r_1,...,r_n) = \bold{v}. It should be clear now how to finish the proof.
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