Another necessary condition for Positive Semidefiniteness?

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Discussion Overview

The discussion revolves around a proposed condition for positive semidefinite (PSD) matrices, specifically whether the maximum absolute value of the off-diagonal elements is less than or equal to the maximum diagonal element. Participants explore the implications of this condition and its potential proof.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that for a hermitian PSD matrix, the maximum absolute value of the entries equals the maximum diagonal entry.
  • The reasoning provided includes the property that all 2x2 principal submatrices of a PSD matrix are also PSD, leading to a relationship between off-diagonal and diagonal elements.
  • Another participant suggests that this condition can be useful in proving whether a matrix is not PSD.
  • A suggestion is made to consider the Schur complement formula to further understand the issues related to PSD matrices.
  • A later reply indicates that the initial proposal is correct, but does not elaborate on the reasoning or implications.

Areas of Agreement / Disagreement

There is no clear consensus on the proposed condition, as some participants express uncertainty and seek clarification, while one participant affirms the correctness of the initial claim without providing detailed justification.

Contextual Notes

The discussion does not resolve the mathematical validity of the proposed condition, and there are references to external sources for further exploration of the topic.

Who May Find This Useful

This discussion may be of interest to those studying linear algebra, particularly in the context of matrix theory and positive semidefinite matrices.

NaturePaper
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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}<br /> a_{ii} & a_{ij} \\<br /> \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:
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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:
Try Schur complement formula to get a feeling for all these issues...
 
@trambolin,
Please let me know whether what I said (guessed) is wrong. Where is the discrepancy?

Regards,
NP
 
@trambolin,
Thanks. Its correct.
 

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