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[Linear Algebra] rank(AT A) = rank(A AT)

  1. Dec 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Does the equation [itex]\text{rank}(A^T A = \text{rank}(A A^T)[/itex] hold for all nxm matrices A? Hint: the previous exercise is useful.


    2. Relevant equations
    [itex] \text{ker}(A) = \text{ker}(A^T A) [/itex]
    [itex] \text{dim}(\text{ker}(A) + \text{rank}(A) = m [/itex]

    3. The attempt at a solution
    The previous exercise it referring to asked to show that [itex]\text{rank}(A) = \text{rank}(A^T A) [/itex] holds for all nxm matrices A.
    Which I did by stating:
    [itex] \text{ker}(A) = \text{ker}(A^T A) [/itex]
    and then taking the dimension of both sides, using the rank-nullity theorem to get:
    [itex] n-\text{rank}(A) = n - \text{rank}(A^T A)[/itex] which makes it clearly true.

    I tried using this result to prove the stated problem like so:
    [itex]\text{rank}(A) = \text{rank}(A^T A) [/itex]
    [itex]\text{rank}(A^T A) = \text{rank}((A^T A)^T A^T A) [/itex]
    [itex]= \text{rank}(A^T A A^T A) [/itex]
    But then I get promptly stuck because I'm not sure what to do with the right side of that. Any advice?
     
  2. jcsd
  3. Dec 11, 2012 #2

    Dick

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    Didn't you also prove rank(A)=rank(A^T)? I.e. column rank equal row rank?
     
    Last edited: Dec 11, 2012
  4. Dec 11, 2012 #3
    Funny you should mention that... That was the problem BEFORE the previous problem, which I still haven't figured out quite yet.

    In any case, I suppose I can use that fact. But I'm not quite sure how it applies since [itex](A^T A)^T = A^T A[/itex]
    So I guess I have these equalities:
    [itex]\text{rank}(A)=\text{rank}(A^T) = \text{rank}(A^T A) [/itex]
    I'm not sure how to rearrange this to get [itex]\text{rank}(A^T A)[/itex]
     
  5. Dec 11, 2012 #4

    Dick

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    But you also have rank(AA^T)=rank(A^T), don't you?
     
  6. Dec 11, 2012 #5
    I do? I guess that what I'm missing, I can't currently see how that arises from the equalities I'm given.

    It would make it trivial from there though, since then I would just say that [itex] \text{rank}(A A^T) = \text{rank}(A) = \text{rank}(A^T A)[/itex]

    I keep wanting to "sub in" [itex]A A^T[/itex] to one of the equalities as a replacement for A, I assume this is the wrong approach?
     
  7. Dec 11, 2012 #6

    Dick

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    Your theorem tells you rank(B^TB)=rank(B) for ANY matrix B. Put B=A^T.
     
  8. Dec 11, 2012 #7
    'doh, thanks a bunch! I should have been able to try that on my own.

    While I'm here, would you mind helping with the other proof I'm stuck on?

    It wanted me to use: [itex] (\text{im} A)^\perp = \text{ker}(A^T) [/itex] to prove [itex]\text{rank}(A)=\text{rank}(A^T)[/itex]

    im is just the column space, and ker is the null space in my textbook.

    I've gotten this far using rank-nullity but I got stonewalled:

    [itex]\text{im } A = \text{ker}(A^T)^\perp [/itex]
    [itex]\text{rank } A = \text{dim}(\text{ker}(A^T)^\perp ) = m - \text{dim } (\text{ker }A)[/itex]

    I feel like this is probably the wrong way to approach this problem but I can't think of another.
     
  9. Dec 11, 2012 #8

    Dick

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    Sorry, it's kind of late here and for some reason this question is making me see double. I'll give you one thing you haven't used yet and that's if V is a subspace of a vector space of dimension n then [itex]dim(V^\perp)+dim(V)=n[/itex]. Hope that helps. I'll have another look in the morning, and if you've gotten it I'll be really happy. I think you can.
     
  10. Dec 12, 2012 #9

    Dick

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    Yeah, it's easy enough. [itex]m-dim(ker(A))=dim(ker(A)^\perp)[/itex]. Got it yet?

    I just edited the above. I had a parenthesis is a bad place.
     
    Last edited: Dec 12, 2012
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