Does $B^{-1}-A^{-1}$ has r positive eigenvalues when A-B has r positive eigenvalues? Assume that A and B are 2 positive definite matrices of $n\times n$. As is known, $A-B>0$ implies $B^{-1}-A^{-1} >0$. That is to say,  $B^{-1}-A^{-1}$  has n positive eigenvalues  when A-B has n positive eigenvalues. 
My question is :Assume that A and B are 2 symmetric matrices of $n\times n$ and B is positive definite and A is invertible.  Does $B^{-1}-A^{-1}$  has r positive eigenvalues  when A-B has r positive eigenvalues ?($r\in \{1,2,\ldots,n\}$)

If $B$ is positive definite, then the following argument applies: note that
$$
B^{-1/2}(A - B)B^{-1/2} = B^{-1/2}AB^{-1/2} - I
$$
is a symmetrc matrix with $r$ positive eigenvalues.  This occurs if and only if $B^{-1/2}AB^{-1/2} - I$ has $r$ eigenvalues that are greater than $1$.  This in turn occurs if and only if $B^{1/2}A^{-1}B^{1/2}$ has $r$ eigenvalues that lie on the interval $(0,1)$. Equivalently, $I - B^{1/2}A^{-1}B^{1/2}$ has $r$ eigenvalues on the interval $(0,1)$. This allows us to deduce that the matrix
$$
B^{-1/2}[I - B^{1/2}A^{-1}B^{1/2}]B^{-1/2} = B^{-1} - A^{-1}
$$
has at least $r$ positive eigenvalues. It could have more than $r$ positive eigenvalues, though.
Concretely, consider
$$
A = \pmatrix{2\\&1/2\\&&-1}, \quad B = I.
$$
Verify that $A - B$ has $1$ positive eigenvalue, but $B^{-1} - A^{-1}$ has $2$ positive eigenvalues.
