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

Question

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.

Answer 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] $.