Why ${(2S)}^{-1} - {(2S)}^{-1}({I + 4S})^{-1} =2\times({I + 4S})^{-1}$ I met an interesting matrix question in a paper. My answer is ${(2S)}^{-1} - {(2S)}^{-1}({I + 4S})^{-1}$, but the answer in the paper is $ 2\times({I + 4S})^{-1}$. When I plugged in some value in $S$, I found they are equal, so why
$${(2S)}^{-1} - {(2S)}^{-1}({I + 4S})^{-1} =2\times({I + 4S})^{-1}$$ 
where $I$ is the identity matrix, and S is a symmetric positive definite matrix.
 A: We can multiply on the right side by $(I+4S)$ and $2S$ on the left to get
$$I+4S-I=2S*2$$
$$4S=4S$$
This last equation is definitely true, and the steps of multiplication and simplifying are reversible as $S$ and $I+4S$ are invertible.
A: We can find strong hints as to why the equation
$(2S)^{-1} - (2S)^{-1}(I + 4S)^{-1} = 2(I + 4S)^{-1} \tag 1$
holds by "unravelling" it; that is, we multiply through by $(2S)(I + 4S)$:
$(2S)(I + 4S)((2S)^{-1} - (2S)^{-1}(I + 4S)^{-1})$
$= (2S)(I + 4S)(2S)^{-1} - (2S)(I + 4S)(2S)^{-1}(I + 4S)^{-1}$
$= (I + 4S) - I = 4S = 2(2S)(I + 4S)(I + 4S)^{-1}; \tag 2$
so if we start with the identity
$(I + 4S) - I = 4S, \tag 3$
and recall that $S$ is given as symmetric positive definite, which is easily seen to imply (and I will leave these details to my readership) that both $2S$ and $I + 4S$ are symmetric positive definite, hence invertible, we obtain
$(2S)^{-1}(I + 4S)^{-1}((4S + I) - I) = (2S)^{-1}(I + 4S)^{-1}(4S), \tag 4$
or
$(2S)^{-1} - (2S)^{-1}(I + 4S)^{-1} = (2S)^{-1}(4S)(I + 4S)^{-1}$
$= \dfrac{4}{2}SS^{-1}(I + 4S)^{-1} = 2(I + 4S)^{-1}, \tag 5$
the desired result.
