A practical way to check if a matrix is positive-definite

Let $A$ be a symmetric $n\times n$ matrix.

I found a method on the web to check if $A$ is positive definite:

$A$ is positive-definite if all the diagonal entries are positive, and each diagonal entry is greater than the sum of the absolute values of all other entries in the corresponding row/column.

I couldn't find a proof for this statement. I also couldn't find a reference in my linear algebra books.

I've a few questions.

1. How do we prove the above statement?

2. Is the following slightly weaker statement true?

A symmetric matrix $A$ is positive-definite if all the diagonal entries are positive, each diagonal entry is greater than or equal to the sum of the absolute values of all other entries in the corresponding row/column, and there exists one diagonal entry which is strictly greater than the sum of the absolute values of all other entries in the corresponding row/column.

Solution 1:

These matrices are called (strictly) diagonally dominant. The standard way to show they are positive definite is with the Gershgorin Circle Theorem. Your weaker condition does not give positive definiteness; a counterexample is $ \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{matrix} \right] $.

Solution 2:

I cannot imagine this is difficult.

Before continuing, let me add the caution that a symmetric matrix can violate your rules and still be positive definite, give me a minute to check the eigenvalues

$$ A_3 \; = \; \left( \begin{array}{rrr} 3 & 2 & 0 \\ 2 & 3 & 2 \\ 0 & 2 & 3 \end{array} \right) , $$

Yes, that is fine, eigenvalues are $3 - \sqrt 8, \; 3, \; 3 + \sqrt 8$

A proof is given here http://planetmath.org/?op=getobj&from=objects&id=7483 as a consequence of Gershgorin's circle theorem. For additional information, see http://en.wikipedia.org/wiki/Diagonally_dominant_matrix and http://mathworld.wolfram.com/DiagonallyDominantMatrix.html or just Google "diagonally dominant symmetric"