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Thread: Doubt about Linear transformation

  1. #1
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    Doubt about Linear transformation

    Hi (excuse me about my english).

    I have a doubt about this linear transformation: f(x) = x - 1

    There is a theorem that says that if the kernel of the transformation is 0; then the transformation is inyective:

    if the ker (f(x)) = 0

    But in this linear transformation the ker (f) = { x = 1 } ; because 1 - 1 = 0

    So the transformation could not be linear, what am I doing wrong?.

    Thank you in advance
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  2. #2
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    Re: Doubt about Linear transformation

    Quote Originally Posted by supertren View Post
    Hi (excuse me about my english).
    I have a doubt about this linear transformation: f(x) = x - 1
    There is a theorem that says that if the kernel of the transformation is 0; then the transformation is injective:
    But in this linear transformation the ker (f) = { x = 1 } ; because 1 - 1 = 0
    So the transformation could not be linear, what am I doing wrong?.
    The theorem is: If $\mathcal{F}$ is a linear transformation is an injection then $\mathcal{Ker(F)}=\{0\}$
    It does not say: If $\mathcal{Ker(F)}=\{0\}$ then $\mathcal{F}$ is a injective linear transformation.
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  3. #3
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    Re: Doubt about Linear transformation

    Quote Originally Posted by Plato View Post
    The theorem is: If $\mathcal{F}$ is a linear transformation is an injection then $\mathcal{Ker(F)}=\{0\}$
    It does not say: If $\mathcal{Ker(F)}=\{0\}$ then $\mathcal{F}$ is a injective linear transformation.
    The theorem is in both directions:

    If $\mathcal{F}$ is a linear transformation is an injection then $\mathcal{Ker(F)}=\{0\}$; and If $\mathcal{Ker(F)}=\{0\}$ then $\mathcal{F}$ is a injective linear transformation
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  4. #4
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    Re: Doubt about Linear transformation

    $f(x)=x-1$, $f:\mathfrak{R}\longrightarrow \mathfrak{R}$,

    is not a linear transformation because $f(x+y)\neq f(x)+f(y)$
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  5. #5
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    Re: Doubt about Linear transformation

    Quote Originally Posted by Idea View Post
    $f(x)=x-1$, $f:\mathfrak{R}\longrightarrow \mathfrak{R}$,

    is not a linear transformation because $f(x+y)\neq f(x)+f(y)$
    OMG, you are right.

    Thanks.
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