Let $\Bbb{S}_{++}^n$ denote the space of symmetric positive definite (SPD) $n\times n$ real matrices, and let $A,B\in\Bbb{S}_{++}^n$.

Is it possible to express the logarithm of $A^{-1}B$ as a difference of the form $P-Q$, where $P,Q$ are $n\times n$ real matrices, not necessarily SPD, but P must depend solely on $A$ and $Q$ must depend solely on $B$, i.e., $$ \log(A^{-1}B)=P-Q. $$

Note that, in genral, $AB\neq BA$.

EDIT: This question is a part of a more general question I asked on MathSE. I am not sure that this is the right approach for attacking my original problem, but I am interested in this one in any case.

Let $\Bbb{S}_{++}^n$ denote the space of symmetric positive definite (SPD) $n\times n$ real matrices, and let $A,B\in\Bbb{S}_{++}^n$.

Is it possible to express the logarithm of $A^{-1}B$ as a difference of the form $P-Q$, where $P,Q$ are $n\times n$ real matrices, not necessarily SPD, but P must depend solely on $A$ and $Q$ must depend solely on $B$, i.e., $$ \log(A^{-1}B)=P-Q. $$

Note that, in genral, $AB\neq BA$.

EDIT: This question is a part of a more general question I asked on MathSE. I am not sure that this is the right approach for attacking my original problem, but I am interested in this one in any case.

Let $\Bbb{S}_{++}^n$ denote the space of symmetric positive definite (SPD) $n\times n$ real matrices, and let $A,B\in\Bbb{S}_{++}^n$.

Is it possible to express the logarithm of $A^{-1}B$ as a difference of the form $P-Q$, where $P,Q$ are $n\times n$ real matrices, not necessarily SPD, but P must depend solely on $A$ and $Q$ must depend solely on $B$, i.e., $$ \log(A^{-1}B)=P-Q. $$

Note that, in genral, $AB\neq BA$.

Let $\Bbb{S}_{++}^n$ denote the space of symmetric positive definite (SPD) $n\times n$ real matrices, and let $A,B\in\Bbb{S}_{++}^n$.

Is it possible to express the logarithm of $A^{-1}B$ as a difference of the form $P-Q$, where $P,Q$ are $n\times n$ real matrices, not necessarily SPD, but P must depend solely on $A$ and $Q$ must depend solely on $B$, i.e., $$ \log(A^{-1}B)=P-Q. $$

Note that, in genral, $AB\neq BA$.

EDIT: This question is a part of a more general question I asked on MathSE. I am not sure that this is the right approach for attacking my original problem, but I am interested in this one in any case.