Prove that $A,B>0$ and $A_{ij}< B_{ij}$ implies $A>0$ 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 $C= c(i,j)= a(i,j) - b(i,j)$, then $C$ is also positive definite. Why or why not?
what i know is $x^T(C)x= x^T (A-B) x = x^T(A)x - x^T(B)x$. Now both terms on right-hand side are greater than $0$ as $A$ and $B$ are positive definite. Also  $a(i,j) < b(i,j)$. But in such case the sign of $x^T(C)x$ being positive or negative will also depend on the value of elements of vector $x$. So, we can't assure that $x^T(C)x < 0$ always or that $C$ is not positive definite.
Am i right about this?
 A: 
Proposition
  If $A=(a_{ij})$, $B=(b_{ij})$ are two positive-definite matrices with $a_{ij}<b_{ij} \forall i,j$, then $A-B$ can only be indefinite or negative-definite or negative semi-definite.  

Demonstration
This strange proposition says only that $A-B$ must take on negative values somewhere, and this is easy to prove: $A-B=(a_{ij}-b_{ij})=(c_{ij})$, so in particular $c_{1,1}<0$, and the form corresponding to $A-B$ contains the monomial $c_{1,1}x_1^2$, and hence must have negative values somewhere. Hence the proposition is proved.  
Examples
The following are some illuminations showing each possibility:
negative-definite: $A=\begin{pmatrix}1\end{pmatrix}, B=\begin{pmatrix}2\end{pmatrix},  A-B=\begin{pmatrix}-1\end{pmatrix}$.
negative semi-definite: $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}, B=\begin{pmatrix}2&1\\1&2\end{pmatrix}, A-B=\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}$.
indefinite: $A=\begin{pmatrix}2&0\\0&2\end{pmatrix}, B=\begin{pmatrix}3&2\\2&3\end{pmatrix}, A-B=\begin{pmatrix}-1&-2\\-2&-1\end{pmatrix}$.
Barring mistakes, and thanks for the attention.  
