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](https://math.stackexchange.com/questions/1334670/show-the-negative-definiteness-of-a-squared-riemannian-metric) 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.



Prove or disprove a matrix logarithm equation
Solution is inspired by the comment of Federico Poloni. Suppose there exist such functions, i.e. $$log(A^{-1}B)=P(A)-Q(B).$$ for all SPD-matrices $A,B$. Putting $A=B$ yields $P(A)=Q(A)$ for all SPD-matrix $A$. Now if the equation above holds we would have $$log(A^{-1}B)+log(B^{-1}C)+log(C^{-1}A)=0.$$
However simply choosing random matrices shows that this indentity does not hold. Here is a matlab code:
d=2;


A=gen\_rand\_spd(d);


B=gen\_rand\_spd(d);


C=gen\_rand\_spd(d);


logm(A^(-1)\*B)+logm(B^(-1)\*C)+logm(C^(-1)\*A)


function [A] = gen\_rand\_spd(d)


A=rand(d);


A=A+A';


A=expm(A);


end


