Another necessary condition for Positive Semidefiniteness?

[NaturePaper]

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


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

Regards,
NaturePaper

 

[NaturePaper]

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

 

[trambolin]

Try Schur complement formula to get a feeling for all these issues...

 

[NaturePaper]

@trambolin,
Please let me know whether what I said (guessed) is wrong. Where is the discrepancy?

Regards,
NP

 

[trambolin]

You can check it for yourself... Horn & Johnson probably contains more than necessary results...

http://en.wikipedia.org/wiki/Positive-definite_matrix

 

[NaturePaper]

@trambolin,
Thanks. Its correct.