Difference of two positive definite matrices Let $A= a(i,j)$ and $B= b(i,j)$ be ($n\times n$) matrices that are positive definite such that $a(i,j) < b(i,j)$.  Let $C= c(i,j)= b(i,j) - a(i,j)$, then $C$ is also positive definite. Why or why not?
 A: Hint: Let $A$ be the $2\times2$ identity matrix, and let $B = A + \pmatrix{ 1 & 1 \\ 1 & 1 }$.
A: Dated, but was interested in this, so thought I'd provide another way of looking at it w/ Gershgorin's theorem: each eigenvalue of any square matrix $A$ lies within one of its Gershgorin disks $(A_{ii}, R_i)$, where $R_i = \sum_{j \neq i} |A_{ij}|$.
Thus if $C = B - A$, each disc is centered at $C_{ii} = B_{ii} - A_{ii} > 0$.  The eigenvalues of $C$ lie in the disks $(b_{ii} - a_{ii},\, \sum_{j\neq i} |a_{ij} - b_{ij}|)$.  
In the case in which $A$ and $B$ are positive semidefinite and $C_{ij} = B_{ij} - A_{ij} \geq 0$, a sufficient condition for $C$ to be positive definite then is that $C_{ii} \geq \sum_{j\neq i} (B_{ij} - A_{ij})$.
A: In general the answer is no. As fuglede has done, we can give an explicit counter-example. 
But it is also interesting to know how to tackle these problem. 
The assumption is that $a(i,j)<b(i,j)$ for all $i,j$. Hence may have $b(i,j)=\varepsilon+a(i,j)$ for some coordinates $i,j$, and $b(i,j)=1+a(i,j)$ for others. 
We have to take the $c(i,j)$ large at corners of the matrix and small at the diagonal, because positive definiteness of a matrix can be thought as a "dominance of the diagonal terms with respect to the non-diagonal ones".
