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Kernel and Image

  1. Apr 19, 2008 #1
    [SOLVED] Kernel and Image

    1. The problem statement, all variables and given/known data
    Ker(A) = Im(B)
    AB = ?

    A is an m x p matrix. B is a p x n matrix.

    2. Relevant equations



    3. The attempt at a solution
    Since Ker(A) is the subset of the domain of B and Im(B) is the subset of the codomain of B, AB = I. I = identity matrix.

    Is this right? It doesn't seem to make sense. There must be a mathematical (symbolic) way to solve for AB, right?
     
  2. jcsd
  3. Apr 19, 2008 #2

    Dick

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    I'm assuming this means B:R^n->R^p. A:R^p->R^m. That way at least Ker(A) and Im(B) live in the same space. Now what do you say AB:R^n->R^m is?
     
    Last edited: Apr 19, 2008
  4. Apr 19, 2008 #3
    Is AB the p x m identity matrix?
     
  5. Apr 19, 2008 #4

    Dick

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    Bad guess. AB is nxm. Consider AB(v). B(v) is in Im(B)=Ker(A). Guess again. Uh, what do you mean by 'identity matrix' anyway. There is no pxm identity matrix. Do you mean the zero matrix?
     
    Last edited: Apr 19, 2008
  6. Apr 19, 2008 #5
    Since Im(B) = Ker(A), does that imply that A and B are invertible? Then if B(v) is in the Ker(A) and Ker(A) = {0}, then AB = 0. Is that right?
     
  7. Apr 20, 2008 #6

    Dick

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    Only one thing you said is right. Why should anything be invertible and why should Ker(A)={0}? If B(v) is in Ker(A) what's A(B(v))? If you think AB=0 then try and put together an argument that will convince me. In your own words.
     
  8. Apr 20, 2008 #7
    Since B(v) is in Ker(A) and T(v) = A * ker(A) = 0, then AB must be zero.
     
  9. Apr 20, 2008 #8

    Dick

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    What's T? I think you MIGHT know what you are trying to say, but that's just a guess. If that's what you are planning to turn in as a solution, I don't think it will work. Can't you put that more clearly?
     
  10. Apr 20, 2008 #9
    The book defines the kernel as a subspace x and T(x) = Ax = 0 and ker(A) = x. T(x) is the linear transformation and A is the matrix. Should I not put the book's equation T(x) = Ax in my solution?

    So in the solution I wrote in my homework, I wrote: BX is in ker(A) and AX = 0, X is in ker(A), therefore AB = 0, is that a good explanation?
     
    Last edited: Apr 20, 2008
  11. Apr 20, 2008 #10

    Dick

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    I don't have your book. What's the linear transformation associated with Bx? Is it also T? What's the linear transformation associated with AB(x). If it's T, I'll scream. You aren't expressing yourself clearly. That's all I'll say.
     
  12. Apr 20, 2008 #11

    Dick

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    It's getting there. Try saying this. For any vector x, Bx is in Im(B). Since Im(B)=Ker(A) then Bx is in Ker(A). If Bx is in Ker(A) then A(Bx)=0. So ABx=0 for all x. So AB=0. How does that sound?
     
  13. Apr 20, 2008 #12
    Thank you for your help! I understand what you mean about my vague explanation. I will try to clarify things a bit on my paper.
     
  14. Apr 20, 2008 #13

    Dick

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    You're welcome. Thanks for letting me scream. Good luck!
     
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