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Thread: Kernal and Image of a Linear Transformation

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    Kernal and Image of a Linear Transformation

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

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

    Show that T is one-to-one if and only if $\displaystyle \{v_1,...,v_n\}$ is independent.
    $\displaystyle T$ is a linear transformation. Therefore, $\displaystyle T$ is one-to-one if and only if $\displaystyle \ker (T) = \{ \bold{0} \}$ if and only if $\displaystyle r_1\bold{v}_1+...+r_n\bold{v_n} = \bold{0} \implies r_1=r_2=...=r_n=0$
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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 $\displaystyle V = span\{v_1,...,v_n\}$
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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 $\displaystyle V = span\{v_1,...,v_n\}$
    A linear transformation $\displaystyle T: \mathbb{R}^n \to V$ is onto iff for any $\displaystyle \bold{v}\in V$ we can find $\displaystyle (r_1,...,r_n)\in \mathbb{R}^n$ so that $\displaystyle T(r_1,...,r_n) = \bold{v}$. It should be clear now how to finish the proof.
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