# Another necessary condition for Positive Semidefiniteness?

1. Jun 26, 2009

### 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
$$\Rightarrow~~\left[\begin{array}{cc} a_{ii} & a_{ij} \\ \bar{a}_{ij} & a_{jj} \end{array}\right]\ge0$$
$$\Rightarrow~~~|a_{ij}|\le \sqrt{a_{ii}a_{jj}}$$
$$\Rightarrow~~\max_{i,j\le n} |a_{ij}|=\max_{i\le n}a_{ii}$$.

Regards,
NaturePaper

Last edited: Jun 27, 2009
2. Jul 3, 2009

### 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

Last edited: Jul 3, 2009
3. Jul 4, 2009

### trambolin

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

4. Jul 5, 2009

### NaturePaper

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

Regards,
NP

5. Jul 8, 2009

6. Apr 4, 2010

### NaturePaper

@trambolin,
Thanks. Its correct.