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Math Help - Idempotent Linear Transformation

  1. #1
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    Idempotent Linear Transformation

    A:V->V is a linear transformation with A^2=A

    Prove that im(A) intersect ker(A) = 0
    I know from a previous exercise that for B, an arbitrary linear transformation, there exists an integer m such that im(B^m) intersect ker(B^m)= 0. I think I can use this and say that A^m=A.

    Prove that V=im(A) direct sum ker(A)
    I know that dim(V)=dim(im(A))+dim(ker(A))
    But I don't know how to use idempotent-ness of A to get set equality. Can I use that the intersection is 0?

    Prove that there is a basis of V such that A with respect to this basis is a diagonal matrix with only 0s and 1s.
    This one I have no idea how to start.
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  2. #2
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    Quote Originally Posted by robeuler View Post
    A:V->V is a linear transformation with A^2=A

    Prove that im(A) intersect ker(A) = 0
    I know from a previous exercise that for B, an arbitrary linear transformation, there exists an integer m such that im(B^m) intersect ker(B^m)= 0. I think I can use this and say that A^m=A.
    In previous section we showed that if \text{im}(A^m) = \text{im}(A^{m+1}) and \ker (A^m) = \ker (A^{m+1}) then \text{im}(A^m) \cap \ker (A^m) = \{ 0 \}. This is a special case with m=1 because A=A^2!
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  3. #3
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    Quote Originally Posted by robeuler View Post
    A:V->V is a linear transformation with A^2=A

    Prove that im(A) intersect ker(A) = 0
    I know from a previous exercise that for B, an arbitrary linear transformation, there exists an integer m such that im(B^m) intersect ker(B^m)= 0. I think I can use this and say that A^m=A.

    Prove that V=im(A) direct sum ker(A)
    I know that dim(V)=dim(im(A))+dim(ker(A))
    But I don't know how to use idempotent-ness of A to get set equality. Can I use that the intersection is 0?

    Prove that there is a basis of V such that A with respect to this basis is a diagonal matrix with only 0s and 1s.
    This one I have no idea how to start.
    Yes the first one I was pretty confident about.
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  4. #4
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    Yes, similar to a group being written as a direct product, I think you only need to quote that dimV=dimim+dimker (G=HK) and im intersect ker = {0} (H intersect K = {0}). Anyone have ideas on the matrix part? All I know is that for that matrix MM=M, so M=I or M is not invertible (singular, detM=0), and then I-M also satisfies (I-M)(I-M)=I-M. Don't see how that gets me anywhere. Help?
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  5. #5
    Senior Member Tinyboss's Avatar
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    Since A^2=A, you have that A restricted to im(A) is the identity. Since you showed that V=im(A)+ker(A), just pick a basis for im(A) and a basis for ker(A), and their union is a basis for V. What does the matrix of A look like under this basis?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Just a remark, since no one else has seemed to say this. The idempotent endomorphisms are precisely the projections.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    Only to add:

    (i)\quad y\in\textrm{Im}A\Rightarrow \exists u \in V:y=Au \Rightarrow Ay=A(Au)=A^2u=Au=y

    (ii)\quad Ay=y\Rightarrow y\in \textrm{Im}A

    So,

    \ker (A-I)=\textrm{Im}A

    Taking into account that

    V=\ker A\oplus \ker (A-I)

    then, A is diagonalizable with diagonal matrix:

    D=\textrm{diag}(0,\ldots,0,1,\ldots,1)

    ( eventually D=0 or D=I ).


    Fernando Revilla

    Edited: Sorry, now I see that Tinyboss had already commented this.
    Last edited by FernandoRevilla; February 6th 2011 at 03:55 AM.
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