Let $b\in \mathbb{R}^d,A\in\mathbb{R}^{d\times d}$ be given.
If $A$ is invertible (with inverse $G$) then $GAw=w$, $w\in \mathbb{R}^d$. What properties of $A,b$ would we need to have to guarantee the existence of a $G$ such that
$$ G(b+Aw)=w ~~~~~~(1)$$
for any $w$? Say we allow $G$ above to be any operator ( not just a matrix ) could one exist for $(1)$ to hold? I mean I can just define the operator like this right?
Given $A$ and $b$, suppose that there exists a matrix $G$ such that, for any $w$:
$$G(b+Aw) = w ~~~~~~(1) .$$
This means that $G$ does not depend on $w$. Rather, it depends on $A$ and $b$ only. We may write:
$$G = G(A, b).$$
Equation (1) is equivalent to:
$$Gb = w - GAw \Rightarrow Gb = (I - GA)w, ~~~~~~(2)$$
where $I$ is the identity matrix.
Suppose that $I-GA$ can be inverted, and let $M$ be the inverse. Again, $M$ depends only on $A$ and $b$, i.e.:
$$M = M(A,b).$$
Now, observe that multiplying both sides of equation (2) by $M$ one gets:
$$MGb = M(I-GA)w \Rightarrow MGb = w. ~~~~~~(3)$$
More precisely, equation (3) reads as:
$$M(A,b)G(A,b)b = w.$$
But this is a contradiction, since all factors on the left hand side does not depend on $w$! Hence, we must conclude that the matrix $G$ does not exist!