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Kernel of the Transpose

  1. Apr 28, 2008 #1
    B= A transpose

    What is the relation between ker(BA) and ker(A)? I was told that they are equal to each other, but I can't figure out why.

    ker(A) => Ax = 0
    ker(BA) => BAx = 0 so that BA is a subset of A. This shows that ker(BA) =0 whenever ker(A) = 0, but how does this also show that they are equal?
  2. jcsd
  3. Apr 28, 2008 #2


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    It might be the case that your shorthand is obscuring things -- try writing things more precisely, and maybe the answer will become more clear.
  4. Apr 29, 2008 #3
    Hi, thanks for your response.

    What do you mean my shorthand? I only said that B = A transpose because I didn't know how to write the superscript T on the forums (is that what you meant?)

    Does showing that ker(A^T*A) is a subset of ker(A) show that they are equal?
  5. Apr 29, 2008 #4
    Let me use Latex for your convenience (you can click on them to learn how to write in case you don't know)...

    [tex]B = A^T[/tex]


    [tex]\ker{(A)} \Rightarrow Ax=0[/tex] and [tex]\ker{(BA)} \Rightarrow BAx = 0[/tex]

    Now plug B in

    [tex]\ker{(A^TA)} \Rightarrow A^TAx = 0[/tex]

    And what do you know about [itex]A^TA[/itex]?
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