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Equivalent conditions for hermitian matrix

  1. Jun 20, 2005 #1
    Let's have hermitian matrix A. Then these three conditions are equivalent:

    1) A is positively definite

    [tex]
    \forall x \in \mathbb{C}^{n} \ {0} : x^{H}Ax > 0
    [/tex]

    2) All eigenvalues of A are positive
    3) There exists regular matrix U such that

    [tex]
    A = U^{H}U
    [/tex]

    Proof:

    [tex]
    2) \Rightarrow 3)
    [/tex]

    A is hermitian => [itex]\exists[/itex] unitary matrix R such that

    [tex]
    A = R^{H}DR
    [/tex]
    where D is diagonal matrix.

    This is what I don't understand - we know that if A is hermitian, there exists R unitary such that

    [tex]
    R^{-1}AR = D \mbox{ diagonal \\}
    [/tex]

    [tex]
    AR = RD
    [/tex]

    [tex]
    A = RDR^{-1}
    [/tex]

    And because R is unitary

    [tex]
    R^{-1} = R^{H}
    [/tex]

    [tex]
    A = RDR^{H}
    [/tex]

    This isn't imho equivalent to

    [tex]
    A = R^{H}DR
    [/tex]

    is it?

    Thank you.
     
    Last edited: Jun 20, 2005
  2. jcsd
  3. Jun 20, 2005 #2

    matt grime

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    I can't quite figure out the best answer, but it as along the general lines of "so what?" Are you claiming R is unique and that it must be written as AR=RD? what is wrong with saying there is a unitary S such that SA=DS? they are equivalent.

    but i can't tell what your problem is; indeed RDR^H and R^HDR are almost always different, but so what?
     
  4. Jun 20, 2005 #3
    I'm not sure whether R is unique. Anyway, I have theorem (the one I wrote into another post):

    A is hermitian => there exists R unitary such that

    [tex]
    R^{-1}AR\mbox{ is diagonal matrix}
    [/tex]

    From this I can get:

    [tex]
    R^{-1}AR = D
    [/tex]

    [tex]
    AR = RD
    [/tex]

    [tex]
    A = RDR^{-1} = RDR^{H}
    [/tex]

    But, the proof of the theorem about equivalent conditions for hermitian matrix says, that

    A is hermitian => there exists R unitary such that

    [tex]
    A = R^{H}DR
    [/tex]


    How can I achieve this? I can't see that..




    Well, that's the point. Why are they equivalent? I just want to use the theorem

    A is hermitian => there exists R unitary such that

    [tex]
    R^{-1}AR\mbox{ is diagonal matrix}
    [/tex]

    which I know is true.
     
  5. Jun 20, 2005 #4

    matt grime

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    You are assuming that the R's in both those theorems are the same; they are not and there is no reason to be. rename the R as S in the first part and we know

    A is hermitian if there is a unitary S such that

    S*AS=E (using * for the H operation)

    where E is diagonal with real entries

    Ie A=SES*

    right?

    now the therem you wnat to prove states as a acondition that there is a unitary R and a diagoanl real D such that A=R*DR

    well, they are equivalent if I let D=E and R=S*

    there is nothing that states the diagonaliztaion of a hermitian matrix is unique (it isn't; reorder the basis).

    you at least see they are equivalent? why must those two R's be the same? Or for that matter, the D?
     
  6. Jun 20, 2005 #5
    Thank you very much Matt, now I'm completely clear about that matter. My fault that I was interpreting the theorem too literally and the substitution R = S* didn't come to my mind. Your answers teach me more that studying from lecture notes :redface:
     
  7. Jun 20, 2005 #6

    matt grime

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    if soemthing states "there is an R" for example, note it is just a label; labelling two things with the same letter here was exactly the problem. lectuer notes won't teach you that since they assume that you won't do that.
     
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