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Positive Definite Matricies

  1. Dec 9, 2012 #1
    on page 261 of this paper by J. Vermeer (http://www.math.technion.ac.il/iic/e..._pp258-283.pdf [Broken]) he writes

    The following assertions are equivalent.
    a) A is similar to a Hermitian matrix
    b) A is similar to a Hermitian matrix via a Hermitian, positive definite matrix
    c) A is similar to A* via a Hermitian, positive definite matrix

    anyway the proof of a)[itex]\Rightarrow[/itex]c) he writes:
    "There exists a V[itex]\in[/itex]Mn(ℂ) such that VAV-1 is Hermitian, i.e. VAV-1=(VAV-1)*=(V*)-1A*V*. We obtain:

    V*VA(V*V)-1=A*

    V*V is the required Hermitian and positive definite matrix."

    My questions is how do we know V*V is positive definite? I know it's Hermitian, i know that V*V has real eigenvalues and I know V*V is unitarily diagonalizable.

    I don't think that V*V is Hermitian is enough right? Does this mean that a matrix B being Hermitian is a sufficient but not necessary condition for B to be positive definite?
     
    Last edited by a moderator: May 6, 2017
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  3. Dec 9, 2012 #2

    quasar987

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    Let e_i be a H-orthonormal diagonalizing basis for V*V. Here H is the standard hermitian product on C^n. The existence of such a basis is equivalent to diagonalizability of V*V by a unitary matrix because the unitary condition is just that the columns are H-orthonormal.
    The ith eigenvalue of V*V is then [itex]H(e_i,V^*Ve_i)=e_i^*V^*Ve_i=e_i^*V^*e_ie_i^*Ve_i=(e_i^*Ve_i)^*(e_i^*Ve_i)=H(e_i^*Ve_i,e_i^*Ve_i)=|e_i^*Ve_i|^2\geq 0[/itex]

    But "=0" is not possible since V*V is invertible. Therefor wrt the basis e_i, the matrix of V*V is diagonal with all nonpositive diagonal entries, so it's positive definite.
     
  4. Dec 9, 2012 #3
    Hi this is really helpful thank you but I have one more question are the ei are the standard basis vectors in ℂn?

    you wrote H(ei, V*Vei)=ei*V*Vei

    =ei*V*eiei*Vei

    I'm a little confused on where this eiei*, this is the matrix with the iith entry being 1 and zeroes everywhere else correct?
     
    Last edited: Dec 9, 2012
  5. Dec 9, 2012 #4

    quasar987

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    Mh! Maybe my argument is flawed. try this much simpler one instead: H(v,V^*Vv) =H(Vv,Vv)=|Vv|^2 for all v. If v is not zero, neither is Vv since V is invertible.
     
  6. Dec 9, 2012 #5
    i thought the first one was nice, umm lemme think about this one for a bit, either way thanks!
     
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