Does $0\prec B \prec A$ imply $A^{-1}B \prec I$ Does $0\prec B \prec A$ imply $A^{-1}B \prec I$? 
I know:  


*

*$A,B$ are of the same size. 

*$A$, $B$ have strictly positive, real eigenvalues.  

*$A^{-1}B$ may not be symmetric since $A^{-1}$ and $B$ may not commute. 

*If $A$ and $B$ are simultaneous diagonalizable, this holds. This is because they have the same orthogonal eigenvectors. By diagonalization, we can prove this. 


But even though $A^{-1}B$ is not symmetric, the definition for $M\prec N$ is that $$x^TMx < x^TNx, \forall x\in \mathbb{R}^n.$$ 
Does this condition hold?  
 A: We are given that $A, B, A-B$ are positive definite, and this also means $A^{-1}, B^{-1}$ are positive-definite. We want to show $I-A^{-1}B$ is positive definite. Note that:
$$I-A^{-1}B=A^{-1}(A-B)$$
Now, this is positive definite if and only if its symmetric part is positive definite, so we will instead look at its symmetric part:
$$\frac 1 2\left(A^{-1}(A-B)+(A-B)A^{-1}\right)$$
Now, based off this forum post, if we choose the following, we will find that the above expression is actually not positive-definite:
$$A^{-1}=\left[\begin{matrix}6 & 0 \\ 0 & 1\end{matrix}\right]$$
$$A-B=\left[\begin{matrix}1/12 & 1/24 \\ 1/24 & 1/24\end{matrix}\right]$$
Now, from these equations, we find:
$$A=\left[\begin{matrix}1/6 & 0 \\ 0 & 2\end{matrix}\right]$$
$$B=\left[\begin{matrix}1/12 & -1/24 \\ -1/24   & 47/24\end{matrix}\right]$$
Here, $A, B, A-B$ are all positive definite, but:
$$A^{-1}B=\left[\begin{matrix} 1/2 & -1/4 \\ -1/48 & 47/48\end{matrix}\right]$$
$$I-A^{-1}B=\left[\begin{matrix}1/2 & 1/4 \\ 1/48 & 1/48\end{matrix}\right]$$
and according to Wolfram Alpha, this matrix $I-A^{-1}B$ is not positive-definite. Thus, these choices of $A$ and $B$ are a counterexample to the theorem your question poses, so $0\prec B \prec A$ does not imply $A^{-1}B \prec I$.
