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Linear Transformation Proof

  1. Jun 1, 2005 #1
    [tex]\ Let T: V \rightarrow W [/tex] be a linear transformation, let [tex]b \in W [/tex]be a fixed vector, and let [tex]x_0 \in V [/tex] be a fixed solution of
    [tex]T(x)=b.[/tex] Prove that a vector [tex]x_1 \in V [/tex]is a solution of [tex] T(x)=b,[/tex] if and only if [tex] x_1 [/tex]is of the form [tex]x_1=x_h +x_0 [/tex]where [tex]x_h \in kerT[/tex]

    I started out by saying that

    [tex] x_i \in X_i[/tex]

    [tex](x_1... x_n) \in \prod [/tex] (where i=1 and h is at the top) [tex]X_0[/tex]

    [tex](x_1... x_n) \in \prod X_i[/tex]

    [tex] x_i \in X_i \rightarrow x_1 [/tex] is not equal to the empty set for all i.

    I am not sure if I am doing this right. I'd appreciate any feedback.
     
    Last edited: Jun 1, 2005
  2. jcsd
  3. Jun 1, 2005 #2

    Hurkyl

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    You're looking for these:

    [tex]\neq \emptyset[/tex]

    [tex]\prod_{i=1}^{h}[/tex]

    [itex]\ker T[/itex]
     
    Last edited: Jun 1, 2005
  4. Jun 1, 2005 #3

    mathwonk

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    to see if you are right, ask yourself if your argument is logically convincing. it is crucial to be able to decide for yourself, if such arguments are correct. i.e. practice playing both roles, argue it then ask if it could possibly be wrong, then answer yourself.
     
  5. Jun 1, 2005 #4
    I think that it's convincing, but sometimes I find it hard to convince myself that it's right because I doubt myself all the time.
     
  6. Jun 1, 2005 #5

    mathwonk

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    well thats the goal to achieve. to reduce your arguments to logic so simple and clear that you can persuade yourself that you must be right. keep practicing.
     
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