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Conditions for an isomorphism

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Let T be defined on F^2 by (x1,x2)T=(w*x1+y*x2, z*x1+v*x2)
    where w,y,z,v are some fixed elements in F.

    (a) Prove that T is a homomorphism of F^2 into itself.
    (b) Find necessary and sufficient conditions on w,y,z,v so that T is an isomorphism.

    3. The attempt at a solution

    I already proved (a).
    Part (b), I'm not sure what it means. For T to be an isomorphism it has to be one-to-one and onto.
    To show one-to-one, I need to show that the kernel is 0.
    Is showing that T is into F^2 the same thing as saying it is onto F^2? If not, what's the difference?
     
    Last edited: Feb 1, 2009
  2. jcsd
  3. Feb 1, 2009 #2

    Mark44

    Staff: Mentor

    Into is different from onto. T is onto F^2 provided that for every (u, v) in F^2 there is some (x1, x2) in F^2 such that T(x1, x2) = (u, v).

    A simple example of an into function is e^x, which maps the real numbers to a subset of the real numbers.

    For the one-to-one part, show that the kernel consists of (0, 0) and nothing else. IOW, if T(x1, x2) = (0, 0), then x1 = 0 and x2 = 0.
     
  4. Feb 1, 2009 #3
    Would it have to be that x1 and x2 equal 0 only? What about the fixed elements in F? Also, the question asks about conditions on the elements of F.
     
  5. Feb 1, 2009 #4

    Mark44

    Staff: Mentor

    Yes, and that's exactly what I said.

    BTW, I didn't notice it earlier, but I think you have copied the definition of the function incorrectly. You have
    I think it should be
    T(x1,x2)=(w*x1+y*x2, z*x1 + v*x2)
    For T to be a map from F^2 onto itself, the image vector has to have two coordinates, not three.

    [tex]\begin{eqnarray}
    T\left (
    \begin{array}{c} \nonumber
    x_1 \\
    x_2
    \end{array}
    \right ) = \left [{\begin{array}{cc}
    w & y \\
    z & v
    \end{array}
    \right ]

    \left ( \begin{array}{c}
    x_1 \\
    x_2
    \end{array}
    \right )\end{equation}[/tex]

    What conditions should you place on the elements of the 2 x 2 matrix to make T one-to-one?
     
  6. Feb 2, 2009 #5
    I don't see what difference the 2x2 matrix would have for T to be one-to-one if we're already saying that x1=x2=0. Is it just that the entries must be scalars? And how can I show that x1=x2=0, i.e. kerT=(0,0), in order to show that T is one-to-one, anyway?
     
  7. Feb 2, 2009 #6

    Mark44

    Staff: Mentor

    There's a connection between the determinant of the matrix of a transformation, and the one-to-one-ness of the transformation, and the dimension of the nullspace of the transformation.

    If you have a linear transformation, it will always be true that T(0, 0) = (0, 0). (I'm assuming the transformation is from a 2D vector space to a 2D vector space in my notation.) The key is whether there are any vectors (x1, x2) that are nonzero, that also map to (0, 0).

    For example, with a different transformation [itex]T: R^2 \rightarrow R^2[/itex], if the matrix of the transformation is [0 1; 0 0], T(1, 0) = (0, 0), so the nullspace does not consist only of {(0, 0)}.
     
  8. Feb 2, 2009 #7
    Oh ok, so the vectors have to be linearly independent? And the determinant has to be zero so that it is invertible?
     
  9. Feb 2, 2009 #8

    Mark44

    Staff: Mentor

    I don't know what that has to do with it.
    No, if the determinant is zero, the matrix is NOT invertible, which means that the transformation isn't one-to-one and doesn't have an inverse.
     
  10. Feb 2, 2009 #9
    Oops, I meant the determinant cannot be zero. So if the determinant is 0 then this means the transformation is one-to-one since it has an inverse?
     
  11. Feb 2, 2009 #10

    Mark44

    Staff: Mentor

    No again. Read what I wrote in my previous post.
     
  12. Feb 2, 2009 #11
    So if the determinant is not 0 then this means the transformation is one-to-one since it has an inverse?
     
  13. Feb 2, 2009 #12

    Mark44

    Staff: Mentor

    Yes.
     
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