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Isomorphism math question

  1. May 16, 2005 #1
    Am I doing this right? I'd appreciate any feedback.

    Let T:U ---> v be an isomorphism. Show that T^-1: V----> U is linear.

    i. T^-1(0) = 0

    ii. T^-1(-V) = -T^-1(V)
    T^-1(-0) = T^-1(0+0)
    = T^-1(0) + T^-1(0)

    T^-1(0) = 0
    T^-1(-V) = T^-1((-1)V)
    =(-1)T^-1(V)
    = -T^-1(V)

    If T[x,y,z] = [x-y, y-z, x+z]
    Then T is one-to-one right?

    How do I show that T is onto?
     
  2. jcsd
  3. May 16, 2005 #2

    mathwonk

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    what is your definition of isomorphism?

    my definition of isomorphism for a linear map T:U-->V is that there exists a linear map S:V-->U such that SoT and ToS are both the identity maps.

    then my version of your statement would be to show that a bijective linear map is an isomorphism, i.e. it has an inverse which is linear.

    i.e. to answer you question, your definition of isomorphism probably assumes T is onto.
     
  4. May 16, 2005 #3
    An isomorphism is something that is both one-ton-one and onto. But, we were also given that if T: W-->V and S:V-->U are linear transformations, then SoT:W-->U is linear.
     
  5. May 16, 2005 #4

    dextercioby

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    [itex] \hat{T} [/itex] linear:

    [tex] \hat{T}(ax+by)=a\hat{T}(x)+b\hat{T}(y) [/tex] (1)

    There exists [itex] \hat{T}^{-1} [/itex] so that [itex] \hat{T}^{-1}\left(\hat{T}(x)\right)=\hat{1}x=x [/tex] (2)

    Therefore

    [tex] \hat{T}^{-1}\left(\hat{T}\left(ax+by\right)\right)=\hat{1}(ax+by)=ax+by [/tex] (3) (by virtue of the definition of isomorphism of vector spaces)

    From (1),it follows that

    [tex] \hat{T}^{-1}\left(\hat{T}\left(ax+by\right)\right)=T^{-1}\left(a\hat{T}(x)+b\hat{T}(y)\right) [/tex] (4)

    Now,again from (2),i write

    [tex] ax+by=a\hat{T}^{-1}\left(\hat{T}(x)\right)+b\hat{T}^{-1}\left(\hat{T}(y)\right) [/tex] (5)

    Comparing (3),(4) & (5),one gets

    [tex] T^{-1}\left(a\hat{T}(x)+b\hat{T}(y)\right)=a\hat{T}^{-1}\left(\hat{T}(x)\right)+b\hat{T}^{-1}\left(\hat{T}(y)\right) [/tex] (6)

    Q.e.d.

    Daniel.
     
  6. May 16, 2005 #5

    mathwonk

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    by the way, you may not appreciate this for a while, but the word isomorphism should not be defined as your source does, but as i did above. the point is that my definition works for all types of maps, linear, continuous, differentiable, group homomorphism, whatever.

    i.e. in many settings a morphism which is one one and onto is still not an isomorphism; e.g. the map taking x to x^3 on the real line does not have a differentiable inverse, but is one one and onto, hence is not a differentiable isomorphism.
     
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