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Showing sets to be equal

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data
    It's given or I've already shown in previous parts of the question:
    [itex]A \in M_{nxn}(F)\\
    A^{2}=I_{n}\\
    F = \mathbb{Q}, \mathbb{R} or \mathbb{C}\\
    ker(L_{I_{n}+A})=E_{-1}(A)[/itex]
    Eigenvalues of A must be [itex]\pm1[/itex]

    Show [itex]im(L_{I_{n}+A})=E_{1}(A)[/itex] where E is the eigenspace for the eigenvalue 1

    (I also need to show that [itex]im(L_{I_{n}-A})=E_{-1}(A)[/itex] but I think that should be simple once I've done one of them)


    2. Relevant equations



    3. The attempt at a solution
    I know that I need to show both sets are contained within the other set so,

    Show [itex]im(L_{I_{n}+A}) \subseteq E_{1}(A)[/itex]
    [itex]y=L_{I_{n}+A}(x)[/itex] Let y be a general element of the image
    [itex]=x+Ax[/itex] By definition of the transformation
    [itex]\Rightarrow A y = A x + A^{2} x[/itex] Multiply through by A
    [itex]= A x + x [/itex] As A2 is the identity element
    [itex]\Rightarrow A y = y \in E_{1}(A)[/itex] As [itex]E_{1}(A) := \{ x | A x = x \}[/itex]

    I've no idea how to show [itex]E_{1}(A) \subseteq im(L_{I_{n}+A})[/itex]
     
  2. jcsd
  3. May 12, 2012 #2

    HallsofIvy

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    You don't seem to have defined "[itex]L_{I_n+ A}[/itex]" in all that.
     
  4. May 12, 2012 #3
    It's never specifically defined in the question but I believe the subscript is the matrix representing the map so [itex]L(x) = (I_{n} + A)x[/itex]
     
  5. May 12, 2012 #4

    micromass

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    Take an eigenvector with eigenvalue 1. So Ay=y.

    You need to find x such that (A+I)x=y.

    What if you take x=y??
     
  6. May 13, 2012 #5
    I don't see how that works, if x=y then Ax=x therefore (A+I)x=Ax+x=2x=y which contradicts itself.
     
    Last edited: May 13, 2012
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