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Prove property of ranks

  1. Jun 8, 2010 #1
    if A is an n x m matrix where n < m I would like to prove that there exists some [tex]\lambda[/tex] such that [tex]rank(A^T A + \lambda I) = m [/tex]

    I know that if two of the columns of [tex]A^T A[/tex] are linearly dependent, they are scalar multiples of each other and by adding some [tex]\lambda[/tex] to two different positions, those colums will become independent but I can't prove it for more than two columns.

    Any tips?
     
    Last edited: Jun 8, 2010
  2. jcsd
  3. Jun 8, 2010 #2
    How many eigen-values does transpose(A)*A has?
     
  4. Jun 8, 2010 #3
    That is not given. A is any n x m matrix.
     
  5. Jun 8, 2010 #4

    Office_Shredder

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    It seems to me that you're taking the pessimistic viewpoint. The rank will be full if the determinant is non-zero. The determinant of [tex]A^T A + \lambda I[/tex] is a polynomial in [tex]\lambda[/tex] , so will have non-zero determinant for all but finitely many [tex]\lambda[/tex].
     
  6. Jun 8, 2010 #5
    A few hints:

    (1) [itex]A^T A + \lambda I[/itex] is full rank iff it's invertible.

    (2) Any strictly Diagonally dominant matrix is invertible.

    What are the diagonal elements of [itex]A^T A + \lambda I[/itex]?
     
  7. Jun 8, 2010 #6
    I got the answer in simpler terms [tex]A^T A[/tex] is symmetric, so it is positive semi-definite and by taking any [tex]\lambda > 0[/tex] the matrix [tex]A^T A + \lambda I[/tex] is positive definite, hence non-singular and invertible.
     
    Last edited: Jun 9, 2010
  8. Jun 8, 2010 #7
    If B is symmetric, then -B is positive-semidefinite? Notwithstanding this, you can choose [itex]\lambda[/itex] such that your matrix is indeed positive definite.
     
  9. Jun 9, 2010 #8
    I meant [tex]A^T A[/tex] is symmetric.
     
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