For $a, b$ two positive constants, we know that \begin{align*} \frac{a-b}{a+b} \end{align*} belongs to the interval (-1, 1).
My question is the following: is there a similar statement for matrices? More precisely,
Let $A, B \in \mathbb{R}^{n \times n}$ be two positive definite symmetric matrices (not necessarily commuting). Is the norm of the matrix \begin{align*} M=(A+B)^{-1}(A-B) \end{align*} strictly less than $1$ (eventually by adding assumptions on $A$ and $B$)?
I tried the following. We can also write this matrix as \begin{align*} M= (I_n + A^{-1}B)^{-1}(I_n - A^{-1}B). \end{align*} We can also write $A^{-1}B = A^{-1/2} (A^{-1/2} B A^{-1/2}) A^{1/2}$. Moreover, since $A^{-1/2} B A^{-1/2}$ is symmetric and positive definite, it may be written as \begin{align*} A^{-1/2} B A^{-1/2} = U^\intercal D U \end{align*} for $U$ unitary and $D$ diagonal positive definite. Injecting these two expressions in the formula for $M$, we obtain \begin{align*} M = A^{-1/2} U^\intercal (I+D)^{-1}(I-D) U A^{1/2}. \end{align*} Then, the norm of $U^\intercal (I+D)^{-1}(I-D) U$ is strictly less than $1$ (because $(I+D)^{-1}(I-D)$ is a diagonal matrix with diagonal entries of the form $(1-d)/(1+d)$ with $d>0$ and $U$ is unitary).
But when is this also true for the norm of $M$? Is there another way to proceed?