Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook