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Another necessary condition for Positive Semidefiniteness?

  1. Jun 26, 2009 #1
    Hi everyone in this sub forum,
    I'm wondering if the following 'rule' (theorem?) is correct:
    For a hermitian Positive Semidefinite (PSD) matrix [tex]A=(a_{ij})[/tex],
    [tex]\max_{i,j\le n} |a_{ij}|=\max_{i\le n}a_{ii}[/tex].


    The reason for this intuition (It may be a well known result, I'm very sorry in this
    case for my poor knowledge) is the following:

    A is PSD [tex]\Rightarrow[/tex] all its [tex]2\times2[/tex] Principal submatrices are PSD
    [tex]\Rightarrow~~\left[\begin{array}{cc}
    a_{ii} & a_{ij} \\
    \bar{a}_{ij} & a_{jj} \end{array}\right]\ge0
    [/tex]
    [tex]\Rightarrow~~~|a_{ij}|\le \sqrt{a_{ii}a_{jj}}[/tex]
    [tex]\Rightarrow~~\max_{i,j\le n} |a_{ij}|=\max_{i\le n}a_{ii}[/tex].

    Regards,
    NaturePaper
     
    Last edited: Jun 27, 2009
  2. jcsd
  3. Jul 3, 2009 #2
    Since I get no reply, I think I'd rather state what it means:

    "For a PSD matrix the largest (consider modulus) entry should necessarily be on a diagonal".

    This sometimes may be a tricky step to prove that a matrix is not PSD.

    Is it write?

    Regards
     
    Last edited: Jul 3, 2009
  4. Jul 4, 2009 #3
    Try Schur complement formula to get a feeling for all these issues...
     
  5. Jul 5, 2009 #4
    @trambolin,
    Please let me know whether what I said (guessed) is wrong. Where is the discrepancy?

    Regards,
    NP
     
  6. Jul 8, 2009 #5
  7. Apr 4, 2010 #6
    @trambolin,
    Thanks. Its correct.
     
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