- #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 [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
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
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