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Proof Help - Rank of the transpose of a Matrix

  1. Apr 24, 2006 #1

    I'm having trouble with a proof regarding the rank of the transpose of a matrix. Here's the question:

    Let A be an m x n matrix of rank r, which is of course less than or equal to min{m,n}. Prove that (A^t)A has the same rank as A.

    Where A^t = the transpose of A.

    I can easily prove that the rank of A^t = rank of A, however I'm havin difficulty with this proof.

    It's urgently needed for an exam coming up in the next day or two, so I appreciate any help you can give me.
  2. jcsd
  3. Apr 24, 2006 #2
    Are you familiar with the dimension theorem? It states for a linear transformation [itex]T:V \rightarrow W[/itex] between vectors spaces V and W,
    nullity(T) + rank(T) = dim(V)

    Note that A, and (A^t)A have the same domain. So one way to prove that they have the same rank, is to show that they have the same null space and then use the result of the dimension theorem.
  4. Apr 25, 2006 #3
    Thanks for your help.

    I've figured out the proof, using the dimension theorem as you suggested. It's not even that difficult once I know what result I'm moving towards.
  5. Oct 5, 2006 #4
    I have the same question as mcintyre_ie. However I till don't understand how to solve it.
    For example, if A is n*m matrix, and of course A^T is m*n matrix, r(A)=m and n>m.
    Than we want to prove (A^T)A has rank m.
    According to nocturnal, V=A, T=(A^T), nullity T is n-mm rank T is m, and rank V=m.
    nullity(T) + rank(T) = dim(V) equal to
    I don't know what it means and how to prove the problem.

    Would you please give me some helps? Thank!
  6. Oct 6, 2006 #5
    In regards to the dimension theorem V is not to be taken as A. V is the domain of the arbitrary linear transformation T. V is a vector space. A is a matrix.

    Let B = (A^t)A. So B is an mxm matrix.
    In my earlier post, I was hinting that you treat both A and B as
    linear transformations.

    Do you know what it means for a matrix to be a linear transformation? If so, what are the domains of A and B. What are their respective ranges?
    Last edited: Oct 6, 2006
  7. Oct 6, 2006 #6
    I think I know how to prove my question.

    I treat (A^T)A as two trsnsformations to prove that the nullity of B is 0. Is it right?
  8. Oct 8, 2006 #7
    No. One of the transformations is left multiplication by the matrix A, the other is left multiplication by the matrix B, which I defined to be (A^T)A.

    I'm going to abandon this approach for the moment and start again. (I think I may be confusing you by calling A and B linear transformations)

    An equivalent statement of the dimension theorem for matrices is that for a matrix A,
    rank(A) + nullity(A) = #columns(A)

    Using the dimension theorem for the matrices A and B we get the following two equations:
    (1) rank(A) + nullity(A) = #columns(A)
    (2) rank(B) + nullity(B) = #columns(B)

    Note that #columns(A) = m = #columns(B), since B = (A^T)A which is an mxm matrix.
    What we'd like to show is that nullity(A) = nullity(B) from which it would follow, via equations (1) and (2), that rank(A) = rank(B).

    To show that nullity(A) = nullity(B), we show that A and B have the same nullspace which amounts to showing:
    [tex] A\hat{x} = \hat{0} \Longleftrightarrow (A^TA)\hat{x} = \hat{0}[/tex]
    Last edited: Oct 8, 2006
  9. Oct 8, 2006 #8
    Thank you for your reply!
    That's what I mean.
    However I also treat the (A^A)A as two transformations to prove the nullity(A)=nullity(B).
    which given an m nonzero vectors belong to space Xm. The first transformation A transform the x nonzero vectors to n nonzero vectors in Xn space because nullity(n)=0. And than the A^T transforms the n vectors to m nonzero vectors in Xm' space because nullity(A^T) is also 0. Thus the transformation of B transforms m nonzero vectors from Xm space to m nonzero vectors in Xm' spce, which means the nullity(B) is 0.

    How do you think if I prove the equation in this way?
  10. Oct 8, 2006 #9
    Im sorry, I can't make sense out of a single thing you said. Rather than go through your post and ask what each statement means, perhaps you could rewrite it, including definitions for any new terms you use (such as Xm). Also look up the definitions of any terms you encounter to make sure you using them correctly, like "nullspace" and "nullity."

    For example, what is Xm? What do you mean treat (A^t)A as two linear transformations, I only see the one, namely (A^t)A, so list the two seperately. What on earth does "nullity(n) = 0" mean?

    Note that you can not just assume that nullity(A)=0. For example take
    [tex] A = \left[
    1 & 0 \\
    0 & 0
    \hat{x} = \left[
    0 \\
    \right] [/tex]

    Then we have

    [tex] \hat{x} \neq \hat{0}[/tex] and [tex]A\hat{x} = \bar{0}[/tex]

    so nullity(A) is not 0.
    Last edited: Oct 8, 2006
  11. Jan 30, 2010 #10
    I had the same question and with your clear explanation I was able to somehow solve it. My question is that to prove ker(A)=ker(AT A) can we set A as say, a 2 by 3 matrix and show iff ([tex] A\hat{x} = \hat{0} \Longleftrightarrow (A^TA)\hat{x} = \hat{0}[/tex])?
    Seems tedious but I did it. The problem is that I don't know whether the assumption I made that”both A and X are ≠0 is correct or not??
    Thanks for the help!

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