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Thread: Lin. dep.

  1. #1
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    Lin. dep.

    Let $\displaystyle f:X\to Y$ be a linear transformation. If $\displaystyle x_1,\cdots , x_n$ are lin. dep., show that $\displaystyle f(x_1),\cdots , f(x_n)$ are lin. dep.

    Since $\displaystyle x_1,\cdots , x_n$ are lin. dep., there exist an $\displaystyle x_i=\sum_{j\neq i}\lambda_j x_j$ where $\displaystyle i=1,\cdots ,n$.

    $\displaystyle f(x_i)=f\left(\sum_{j\neq i}\lambda_j x_j\right)=\sum_{j\neq i}\lambda_j f(x_j)$

    $\displaystyle f(x_i)=\sum_{j\neq i}\lambda_j f(x_j)\Rightarrow 0=\left[\sum_{j\neq i}\lambda_j f(x_j)\right]-f(x_i)$

    Which is a nontrivial sol. since not all $\displaystyle \lambda_i$ are equal to zero.

    Correct?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Lin. dep.

    Quote Originally Posted by dwsmith View Post
    Correct?
    Yes, it is correct.
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