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Linearly independence question

  1. Nov 3, 2013 #1
    Say I have a matrix ##A## that has linearly independent columns. Then clearly ##A^T## has lin. indep. rows. So what can we say about ##A^TA##? Specifically, is there anything we can say about the rows/columns of ##A^TA##? I'm thinking there has to be some sort of relation but I don't know what that is (if there is indeed any).
     
  2. jcsd
  3. Nov 3, 2013 #2
    We have [tex]\textrm{rank}(A) = \textrm{rank}(A^T A)[/tex] So the number of linear independent columns/rows of ##A## is the same as the number of linear independent columns/rows of ##A^T A##.
     
  4. Nov 3, 2013 #3
    Isn't ##\textrm{rank}(A) = \textrm{rank}(A^TA)## only true if ##A## is symmetric? Also, I forgot to include that ##A## is not necessarily a square matrix. Let's have ##A## be an ##n## x ##k## matrix. Does your conclusion still follow with these new conditions?
     
  5. Nov 3, 2013 #4
    Yes, it is true in general. Indeed, by rank-nullity is suffices to show that the nullity of ##A## equals the nullity of ##A^T A##.

    But take ##Ax = 0##, then obviously ##A^T A x = 0##.
    Conversely, if ##A^T A x = 0##, then ##x^T A^T A x##. But then ##|Ax| = 0##. Thus ##Ax= 0##.

    So the nullspace of ##A## equals the nullspace of ##A^T A##.
     
  6. Nov 3, 2013 #5
    I was under the impression that ##\textrm{rank}(A) = \textrm{rank}(A^TA)## is only true if ##A## is symmetric, but it appears you are right, and Wikipedia confirms it. It is indeed true in general for any ##A##, so I guess I misread it somewhere. Thanks for the help!
     
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