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Prove that rank(A) = rank(A^T)

  1. Apr 22, 2017 #1

    hotvette

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    1. The problem statement, all variables and given/known data
    Use the below relation to prove that [itex]rank(A) = rank(A^T)[/itex]

    2. Relevant equations
    [tex](Im A)^c = ker(A^T)[/tex] where c is the orthogonal compliment.

    3. The attempt at a solution
    Let A be an n x m matrix with [itex](n \ge m)[/itex] and [itex]rank(A) = k[/itex]. Then, [itex]rank(A^c) = n-k[/itex]. From the given relation, [itex]rank(ker A^T)[/itex] must also equal [itex]n-k[/itex]. Thus, [itex]rank(A^T) = n - (n-k) = k = rank(A)[/itex].

    Is this a valid argument?
     
  2. jcsd
  3. Apr 22, 2017 #2

    andrewkirk

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    It looks like it's probably correct, but the explanation is a bit confusing because it uses operators in what appear to be non-standard ways. For instance if c denotes orthogonal complement, that is a function whose domain is a vector space, so it cannot be applied to a linear operator, which is what the symbol ##A^c## suggests. Also, I have not seen the rank operator applied to a space rather than to a linear operator, which is what ##rank(ker\ A^T)## suggests. Perhaps it means 'dim'(ension)?
     
  4. Apr 25, 2017 #3

    hotvette

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    Thanks, I agree that my explanation used sloppy terminology. Maybe this is better: Let [itex]A[/itex] be an n x m matrix with [itex]n \ge m[/itex] and [itex]rank=k[/itex]. Then [itex]dim((Im A)^c) = m-k = dim(ker(A^T))[/itex]. Also, [itex]dim(ker(A^T)^c)=m-(m-k)=k[/itex]. But, [itex](ker(A^T))^c = Im(A^T)[/itex]. Therefore, [itex]dim(Im(A^T))=k=dim(Im A)[/itex], and thus [itex]rank(A^T)=rank(A)[/itex]
     
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