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Covariance matrix

  1. Jul 7, 2010 #1
    I've been reading everywhere, including wikipedia, and I can't seem to find a prove to the fact that the covariance matrix of a complex random vector is Hermitian positive definitive. Why is it definitive and not just simple semi-definitive like any other covariance matrix?
    Wikipedia just states this and never provides with a prove. Some people might say that it follows from the definition and that a prove is trivial, but I just can't seem to find why. Even if the prove is trivial (and it's eluding me) can somebody please demonstrate why?

    Also, if this is true, does it also apply for crosscovariance matrix? (between two different complex random vectors)
     
  2. jcsd
  3. Jul 8, 2010 #2
    Every real symmetric matrix is Hermitian and there is a complex to real isomorphism for Gaussian RVs. I believe this arises from the definition of Hermitian matrices. I'm not sure if the isomorphism holds for the covariance matrix of non-Gaussian RVs.

    http://www.ee.ic.ac.uk/hp/staff/dmb/matrix/expect.html
     
    Last edited: Jul 8, 2010
  4. Jul 9, 2010 #3
    Thanks for the reply, in that post it states that the matrix is positive semi-definite. And the reference I found somewhere else said that the covariance matrix is always positive-definite. Thanks a lot for your reply!!!
     
  5. Jul 9, 2010 #4
    Sorry, for some reason the wrong link came up. I've got one that answers your question re covariance matrices.

    http://www.riskglossary.com/link/positive_definite_matrix.htm
     
    Last edited by a moderator: Apr 25, 2017
  6. Jul 10, 2010 #5
    There are no constraints on the elements of a crosscovariance matrix, so it is clearly not necessarily positive semidefinite (or even square in shape!)
     
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