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I Proof that the general solution of a linear equation is....

  1. Mar 19, 2016 #1
    any particular solution plus the general solution to the homogeneous equation.

    I'm having difficuilty understanding this proof from my lecture notes

    : Let T : V → W be a linear transformation. Let w ∈ W and suppose T(u0) = w

    T(v) = 0. where v ∈ V (the kernel )

    to prove:
    T(u) = w, where u = u0 + v

    proof: (from my lecture notes)

    First, we show that every vector of the form u = u0 + v
    where T(u0) = w and T(v) = 0, satisfies T(u) = w:
    T(u) = T(u0 + v) = T(u0) + T(v) = w + 0 = w

    Now we show that every solution looks like this.
    Suppose that T(u) = w. Let v = u − u0.
    Then u0 + v = u0 + (u − u0) = u and v is in the kernel of
    T: T(v) = T(u − u0) = T(u) − T(u0) = w − w = 0

    I can follow the first part of the proof. The second part is where I'm having difficulty. I'm not sure how the last three lines show that every solution looks like u = u0 + w, particularly the second last line. Is that not showing what we were already given?
    Any help would be appreciated
  2. jcsd
  3. Mar 19, 2016 #2


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    "Suppose that T(u) = w. Let v = u - u0"
    "Suppose that T(y) = w. Let v = y - u0 "
    and now it no longer shows what we were already given !
  4. Mar 19, 2016 #3


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    This is the way that makes sense to me:

    Assume T(a)=w =T(b). Then T(a)-T(b)=T(a-b)=0 , so a-b ## \in K ## ; K :=Kernel T.

    Then a-b=k, so a =b+K. Then T(b)=w iff b=a-K . EDIT: BTW this is related to one of the
    isomorphism theorems stating that if T: V-->W is a homeomorphism (a linear map, in our case)

    with kernel K , then T: V/K --> T(W) is an isomorphism. Moding out by the kernel just collapses
    elements with the same preimage, so you get a linear bijection between V/K and the image.
    Last edited: Mar 19, 2016
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