TITLE: A practical way to check if a matrix is positive-definite
QUESTION [28 upvotes]: 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.

How do we prove the above statement?
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.

REPLY [5 votes]: This is in part a rehash of Noah Stein answer, without too much clutter. More importantly, it gives you an algorithmic way to determine what type of matrix you have.
It's easy if you follow Cholesky method in here. 
A squared matrix is positive definite if it is symmetric (!) and $x^TAx>0$ for any $x\neq0$. Then by Cholesky decomposition theorem $A$ can be decomposed in exactly one way into a product
$$
A = R^TR
$$
where $R$ is upper triangular and $r_{ii}>0$.
If this is true, then (see the reference!), the diagonal elements of $R$ must fulfill
$$
r_{ii} = +\sqrt{a_{ii}-\sum_{k=1}^{i-1}r_{ki}^2}\; j=i+1,\dots,n,
$$
where $n$ is the real vector space dimension.
If at some point this square root is either complex or zero, then $A$ cannot be positive definite.
Nice and neat! You just have to squared, take the difference and the square root to find if $A$ is positive. It is even easy to program this method even for arbitrary $n$.