Determine if a matrix if positive definite

[onako]

Given a symmetric matrix
$$A=\left(\begin{array}{ccccc}<br /> \sum a_{1s} & & & & \\<br /> & \ddots & & a_{ij} \\<br /> & & \ddots & & \\<br /> &a_{ij} & & \ddots & \\<br /> & & & & \sum w_{as}<br /> \end{array}\right) \in\mathbb{R}^{n\times n},$$
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,
$$x^TAx>0$$
for some non-zero vector x.

 

[AlephZero]

The http://en.wikipedia.org/wiki/Gershgorin_circle_theorem shows there are no negative eigenvalues, but it doesn't exclude the possibiltiy of zero eigenvalues (i.e. a singular matrix).

In fact the matrix
$$\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}$$
is singular, and therefore not positive definite.

 

[onako]

Thanks for providing the example.
I guess the author of the book stating the above positive-definiteness on the given matrix type
somehow misinterpreted it.

 

[morphism]

Thanks for providing the example.
I guess the author of the book stating the above positive-definiteness on the given matrix type
somehow misinterpreted it.

Actually, unless I'm mistaken, the matrix is invertible (hence positive definite) if n>3.

 

[Jim Kata]

This is true because the matrix is diagonally dominant. There is a theorem that says a Hermitian diagonally dominant matrix with real nonnegative diagonal entries is positive semidefinite. A proof of this is found here http://planetmath.org/?op=getobj&from=objects&id=7483

 

[onako]

It can be shown that the inequality holds for n>3, but not in general case, as is observed above.