# 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.

• What is the original question in the paper?
– paf
Jul 2, 2018 at 18:16
• It is a long proof, and $S$ is the covariance matrix. Jul 2, 2018 at 18:22

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.

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.