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Linear Transformation - Onto

  1. Apr 18, 2004 #1
    Linear Transformation -- Onto

    I'm having trouble with the first part of the following problem:

    Let T be a linear transformation from an n-dimensional space V into an m-dimensional space W.

    a) If m>n, show that T cannot be a mapping from V onto W.

    b) if m<n, show that T cannot be one-to-one.

    Part b) I can see. I think. T(v) = Av = w The matrix A will have more columns than rows (more unknowns than equations), so there will be infinitely solutions (more than one mapping from a v in V to a w in W).

    I'm stumped by part a). I'm not seeing how m>n guarantees that there are w 's in W that aren't part of R(T).

    A nudge in the right direction would be greatly appreciated.

    Thanks.
     
    Last edited: Apr 18, 2004
  2. jcsd
  3. Apr 18, 2004 #2

    Hurkyl

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    Why doesn't the same approach work?
     
  4. Apr 18, 2004 #3
    I'm wondering the same thing, so there must be something that I'm not seeing.

    I'll think about it some more.

    Thanks.
     
  5. Apr 19, 2004 #4
    Well, for a linear map: [tex]T:V^n \rightarrow W^m[/tex] where [tex]n<m[/tex]

    There is a useful formula which describes subspaces of V and W in terms of conditions they satisfy with respect to T. Then have a look at the dimension of these subspaces, the dimension of V, and the dimension of W. You should be able to establish that there are certain elements in W that aren't the image of any element in V. Hint: Consider a basis of W.
     
    Last edited: Apr 19, 2004
  6. Apr 19, 2004 #5

    NateTG

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    Hint: Do you think that T inverse is defined for all of W?
     
  7. Apr 19, 2004 #6
    OK, let's see what I've got so far:

    A basis of W would consist of 3 elements, A basis of V would consist of 2 elements.

    T(x + y ) = T(x ) + T(y )
    T(kx ) = kT(x )

    I also have the equation dim(ker(T)) + dim(R(T)) = dim(R2) = 2

    Anyway you look at it dim(R(T)) <= 2. This means that any other element in R(T) is a linear combination of these two elements. But W has a basis of 3 elements meaning that R(T) could not possibly contain at least one of W's basis elements.

    I think this makes sense finally. I'll think about it some more just to make sure it's solid.

    Thanks very much for the hints.

    discoverer02
     
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