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A square matrix A with ker(A^2)= ker (A^3)

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    a square matrix A with ker(A^2)= ker (A^3), is ker(A^3)= ker (A^4),verify...


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2011 #2

    micromass

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    What did you try already??
     
  4. Sep 20, 2011 #3
    it means that A^2*vector x= 0 and A*x=0 has same result,then I really confused how to do the next step
     
  5. Sep 20, 2011 #4

    micromass

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    No, it means that

    [tex]A^2x=0~\Leftrightarrow~A^3x=0[/tex]

    You need to prove that

    [tex]A^3x=0~\Leftrightarrow~A^4x=0[/tex]
     
  6. Sep 20, 2011 #5
    emm...., to the difinition it says that ker(A) is T(x)=A(x)=0
     
  7. Sep 20, 2011 #6

    micromass

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    Yes, but you're not working with ker(A) here, but with ker(A2).
     
  8. Sep 20, 2011 #7
    should define a matrix A and write A^2 as dot product of A & A,and also to A^3...it seems to be so complex, or use the block to devide the matrix into some small matrix?
     
  9. Sep 20, 2011 #8

    micromass

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    Do you understand my Post 4??
     
  10. Sep 20, 2011 #9
    so times A on both side of the equation?
     
  11. Sep 20, 2011 #10

    micromass

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    No, you need to prove two things:

    If [itex]A^3x=0[/itex], then [itex]A^4x=0[/itex]. This can indeed be accomplished by multiplying sides with A.

    But you also need to prove that if [itex]A^4x=0[/itex], then [itex]A^3x=0[/itex].
     
  12. Sep 26, 2011 #11
    so Ker (A^2)=0 can lead to Ker(A^4)=0,then?
     
  13. Sep 26, 2011 #12

    HallsofIvy

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    NO ONE has said that Ker(A^2)= 0 so I do not understand why you are asking this question. You seem, frankly, to have no idea what the question is saying. ker(A^2)= ker(A^3) means, as micromass said, "A^2x= 0 if and only if A^3x= 0". You want to use that to prove "A^3x= 0 if and only if A^4x= 0". As micromass said, the first part is easy: if A^3x= 0 then, applying A to both sides, A^4x= A0= 0 which proves that ker(A^3) is a subset of ker(A^4). You still need to prove the other way: if A^4x= 0, then A^3= 0. You cannot just multiply by A^{-1} because you have no reason to think that A is invertible. But notice that ker(A^2)= ker(A^3) has not yet been used so it might help to note that A^4x= A^2(A^2x).
     
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