I was reading on mathworks (http://mathworld.wolfram.com/PositiveDefiniteMatrix.html) that a Hermitian positive definite matrix $[a_{ij}]$ necessarily satisfies \begin{equation*} a_{ii} + a_{jj} > 2 |\mathcal{R}[a_{ij}]| \end{equation*} Can someone tell me where this comes from please?
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Consider the vector with $1$s in positions $i$ and $j$ and zero elsewhere. – Angina Seng Aug 30 '17 at 14:44
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If $A$ is positive definite, then so is every principal submatrix. Thus, the matrix $$ A[\{i,j\}] = \pmatrix{a_{ii} & a_{ij}\\a_{ji} & a_{jj}} $$ is positive definite (and we note that $a_{ji} = \overline{a_{ij}}$).
With that in mind, take $x = (1,z)^T$ and note that $$ x^*A[\{i,j\}]x = a_{ii} + |z|^2 a_{jj} + za_{ij} + \overline{za_{ij}} = a_{ii} + |z|^2 a_{jj} + 2 \Re [za_{ij}] > 0 $$ Now, consider this inequality for an appropriate $z$ satisfying $|z| = 1$.
Ben Grossmann
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