How prove this matrix $B^{-1}-A^{-1}$ is positive-semidefinite matrix,if $A-B$ is positive matrix

Question

Question:

Let $A,B$ be positive $n\times n$ matrices, and assume that $A-B$ is also a positive definite matrix.

Show that $$B^{-1}-A^{-1}$$ is a positive definite matrix too.

My idea: since $A,B$ be positive matrix,and then exist non-singular matrices $P,Q$,such $$A=P^{-1}\operatorname{diag}{(a_{1},a_{2},\cdots,a_{n})}P$$ $$B=Q^{-1}\operatorname{diag}{(b_{1},b_{2},\cdots,b_{n})}Q$$ where $a_{i},b_{i}>0$

Answer 1

As $A$ and $B$ positive definite they have positive definite square roots: $$ A=A_1^2,\,\, B=B_1^2. $$ [If $A=U^*\mathrm{diag}(d_1,\ldots,d_2)U$, then $A_1=U^*\mathrm{diag}(d_1^{1/2},\ldots,d_2^{1/2})U$.]

Clearly$^*$, $$ A-B\ge 0 \Longleftrightarrow B^{-1}_1AB_1^{-1} \ge I, $$ where $I$ is the unit matrix. Also$^{**}$, $$ B^{-1}_1AB_1^{-1} \ge I\Longleftrightarrow I \ge B_1A^{-1}B_1 \Longleftrightarrow B^{-1} \ge A^{-1}. $$

$^*$More specifically, if $(x,Ax)\ge (x,Bx)$, then $(B_1^{-1}x,AB_1^{-1}x)\ge (B_1^{-1}x,BB_1^{-1}x)$ or $(B_1^{-1}x,AB_1^{-1}x)\ge (B_1^{-1}x,B_1x)$, for all $x$. But $(B_1^{-1}x,B_1x)=(x,B_1^{-1}B_1x)=(x,x)$ and $(B_1^{-1}x,AB_1^{-1}x)=(x,B_1^{-1}AB_1^{-1}x)$.

$^{**}$If $A\ge I$, then $(x,Ax)\ge(x,x)$, and hence $$ (x,x)=(A_1^{-1}x,AA_1^{-1}x)\ge(A_1^{-1}x,A_1^{-1}x)=(x,A^{-1}x). $$