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Thread: ker f

  1. #1
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    ker f

    Let $\displaystyle f:X\to Y$ be a lin transformation. f is monic iff $\displaystyle \text{ker}f=\{0\}$

    I have shown $\displaystyle (\Rightarrow)$

    I am struggling with $\displaystyle (\Leftarrow)$

    Let $\displaystyle x\in\text{ker}f$. Then $\displaystyle f(x)=\{0\}$.

    Now what?
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  2. #2
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    Re: ker f

    If I am not mistaken, the fact that f is monic means that $\displaystyle f\circ g_1=f\circ g_2$ implies $\displaystyle g_1=g_2$ for all $\displaystyle g_1,g_2$. Well, let $\displaystyle f(g_1(x))=f(g_2(x))$ for all x. Use the fact that f is linear and that its kernel is {0}.
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  3. #3
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    Re: ker f

    Quote Originally Posted by dwsmith View Post
    Let $\displaystyle f:X\to Y$ be a lin transformation. f is monic iff $\displaystyle \text{ker}f=\{0\}$

    I have shown $\displaystyle (\Rightarrow)$

    I am struggling with $\displaystyle (\Leftarrow)$

    Let $\displaystyle x\in\text{ker}f$. Then $\displaystyle f(x)=\{0\}$.

    Now what?
    You have to show that $\displaystyle f$ is monic.

    So, you have to show that $\displaystyle f(x) = f(y)$ implies that $\displaystyle x = y$.

    Now $\displaystyle f(x) = f(y) \Rightarrow f(x-y) =0 \Rightarrow x-y \in Ker(f) \Rightarrow x-y = 0 \Rightarrow x=y$.
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