- #1

onako

- 86

- 0

[tex]A=\left(\begin{array}{ccccc}

\sum a_{1s} & & & & \\

& \ddots & & a_{ij} \\

& & \ddots & & \\

&a_{ij} & & \ddots & \\

& & & & \sum w_{as}

\end{array}\right) \in\mathbb{R}^{n\times n},

[/tex]

with strictly positive entries a_{ij}, and with the diagonal entries being sum of off-diagonal entries residing

in the corresponding row/column, how to proceed with the proof for A being positive definite,

[tex]

x^TAx>0

[/tex]

for some non-zero vector x.