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Math Help - Positive-definite Hermitian matrix

  1. #1
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    Positive-definite Hermitian matrix

    Hi,

    For any Hermitian matrix T,

    z* T z is real and is >= 0. The * denotes conjugate transpose.

    I understand that if any of the eigenvalues of T are equal to zero, then T is positive semi-definite (z* T z >= 0).

    What I don't get is why T is positive definite (z* T z > 0) iff all of it's eigen values are > 0; even if all eigenvalues are > 0, couldn't there still be a z which leads to semi-definiteness?

    Thanks.
    Last edited by algorithm; April 4th 2012 at 12:27 PM.
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  2. #2
    Super Member girdav's Avatar
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    Re: Positive-definite Hermitian matrix

    No, since T is Hermitian it is diagonalizable, and if z^*Tz=0 then z=0.
    Thanks from algorithm
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  3. #3
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    Re: Positive-definite Hermitian matrix

    Thanks.

    Could you please relate the diagonalizability to the part that if z* T z = 0 then z = 0?
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  4. #4
    Super Member girdav's Avatar
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    Re: Positive-definite Hermitian matrix

    We can write T=P^*DP where P is unitary and D diagonal, containing the eigenvalues of T. If z^*Tz=0 then (Pz)^*T(Pz)=0, but writing v=Pz we have v^*Tv\geq \min_{1\leq j\leq d}\lambda_j\lVert v\rVert so v=0 and z=0.
    Thanks from algorithm
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