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Linera Transformation Questions

  1. Jul 21, 2008 #1
    Hey everyone,

    Let T: W -> V be a linear transformation and T(w_1) = v_1 for some w_1 \in W and v_1 \in V. Now set S = {w \in W | T(w) = T(w_1) = v_1}. Prove that S = w_1 + Kernel(T) = {w_1 + a | a \in Ker(T)}.

    Let w \in S. T(w_1) = T(w) = T(w) + T(a) = T(w_1) + T(a) = T(w_1 + a) = v_1, where a \in Ker(T).

    Thus

    S_1 = {w \in W | T(w) = T(w_1) = v_1}; S_2 = {w_1 + a | a \in Ker(t)}

    S_1 = S_2 = S and we're done. Correct?
     
    Last edited: Jul 21, 2008
  2. jcsd
  3. Jul 21, 2008 #2
    Hmm....

    All elements of S = S_1 constitute elements whose images are v_1, and the elements of S_2 constitute elements whose images are v_1. Now how can I prove they're no preimages in S_1 that aren't present in S_2 and vice versa?
     
  4. Jul 21, 2008 #3

    Defennder

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    Homework Helper

    This looks ok, though the steps appear to be out of order.

    You mean how do you prove there isn't any [tex]\vec{u} \in W \ | \ T(\vec{u}) \neq \vec{v_1}[/tex]? Didn't you just do that with the above? Note that by definition all such [tex]\vec{u} \ \mbox{can be expressed as} \ \vec{w_1} + \vec{a} [/tex]. What does that tell you?
     
  5. Jul 21, 2008 #4

    HallsofIvy

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    If x is any member of S then T(x)= T(w_1)= w by definition of S. Now, let a= x- w_1 so that x= w_1+ a. It is easy to prove T(a)= 0 (do that yourself) and so x is in S_2.

    Conversely, if x is any member of S_2, then x= w_1+ a by definition of S_2. But then T(x)= T(w_1+ a)= T(w_1)+ T(a)= ? (and so x is in S).
     
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