- #1
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 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<br />
\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
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<br />
\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
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