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Condition on Leading principle minors of a symmetric Positive semidefinite(PSD) matri

  1. Apr 4, 2010 #1
    Hi everyone,
    This is related to my previous https://www.physicsforums.com/showthread.php?t=392069"

    Let [tex] A=(a_{ij}) [/tex] be a symmetric (i.e., over reals) PSD matrix with the following conditions on Leading Principle Minors (determinant of the submatrix consisting of first i rows and i columns) [tex] A_{ii}[/tex]:

    [tex] A_{11}\ge0,~ A_{22}=A_{44}= A_{66}=A_{77}=A_{88}=detA=0 [/tex]

    Now the question is can I say (from the above information) that [tex] A_{33}=A_{55}=0 ?[/tex] From "Matrix Analysis" by Horn and Johnson, I guess the Interlacing Inequlity may be useful...but I don't know much about it. Any help, please.

    As usual, will it still valid if I assume A to be hermitian (i.e., over complex) than being symmetric?

    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Apr 4, 2010 #2
    Re: Condition on Leading principle minors of a symmetric Positive semidefinite(PSD) m

    Oh, I got the answer. It is indeed yes.

    The result follows directly from Theorem 4.3.8 (page-185) of the book I mentioned above. It is a consequence of the "Interlacing inequality" as I guessed. Below is a brief sketch:

    By our assumption, [tex]A_{22}[/tex] must have an eigenvalue 0 and hence by the interlacing property, the least eigenvalue (which can not be negative as it is PSD) of [tex] A_{33}\mbox{~is} \le0[/tex]. Thus follows.

    The result remains valid if A be hermitian.

    More generally, we can say that for a hermitian PSD matrix [tex]A_{KK}=0~\Rightarrow~ A_{MM}=0~\forall M>K[/tex]
    Last edited: Apr 4, 2010
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