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Isomorphism: subspace to subspace?

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    We're looking at a mapping from P2 (polynomials of degree two or less) to M2(R) (the set of 2x2 real matrices). The nuance here is that the transformation into the matricies is such that its basis consists of only three independent matrices, making its dimension 3. This means that our transformation maps from P2 (dim = 3) to M2(r) (dim = 3 in this case)
    Can a mapping to a subspace make the transformation an isomorphism?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 26, 2011 #2

    micromass

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    Yes, this mapping might be an isomorphism on the sibspace. Do you know whether it is surjective?? This is already enough for showing it an isomorphism.

    Indeed: if V and W both have equal dimension then a map [itex]T:V\rightarrow W[/itex] is an isomorphism if and only if it is surjective and linear (or injective and linear). This is the alternative theorem.
     
  4. Nov 26, 2011 #3
    The transformation itself is
    P2 →M2(ℝ)
    T(ax2 + bx + c) → Matrix(a11 = -b-a, a12 = 0, a21 = 3c-a, a22 = -2b)
     
    Last edited: Nov 26, 2011
  5. Nov 26, 2011 #4

    micromass

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    Yes, so is the range three dimensional??
     
  6. Nov 26, 2011 #5
    Yes, the range itself is a three dimensional subspace of the four dimension space of 2x2 real matrices. I'm just not sure if it's an isomorphism.
     
  7. Nov 27, 2011 #6

    micromass

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    What is left to check for it to be an isomorphism??
     
  8. Nov 27, 2011 #7
    Well we need to know Dim(V) = Dim(W), but what I'm curious about is if we take the dimension from the general mapping P2 to M2(R) or from the actual transformation.
    We're not actually onto if we consider the general mapping, but we're onto if we consider the subspace mapping.

    The other way to check would be finding the matrix of the transformation and seeing if it's invertible, but I've had no luck with that.
     
  9. Nov 27, 2011 #8
    I'm still at a loss with this problem :/
     
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