If A and B are non-singular and n-square matrices. Show that $(I+BA)^{-1}=I-(((B^{-1})+A)^{-1})A$

Question

A and B are non-singular n-square matrices. Show that

$(I+BA)^{-1}=I-(((B^{-1})+A)^{-1})A$

Answer 1

Assuming that $(I+BA)^{-1}$ exists,

$$\begin{align}I-((B^{-1}+A)^{-1})A&=I-([B^{-1}(I+BA)]^{-1})A\\&=I-(I+BA)^{-1}BA\\&=I-(I+BA)^{-1}(I+BA-I)\\&=I-(I+BA)^{-1}(I+BA)+(I+BA)^{-1}\\&=I-I+(I+BA)^{-1}\\&=(I+BA)^{-1}\end{align}$$

Answer 2

As a hint, the statement is equivalent to:

$$(I+BA)[I-((B^{-1}+A)^{-1})A]=I$$

Can you check if that is a true statement?

Answer 3

Hint. Substitute $C=B^{-1}+A$ and simplify $$(I+AB)(I-(B^{-1}+A)^{-1}A)\ .$$

Note that your claim is not quite true: if $A=-B^{-1}$ then $I+AB=0$ and this certainly has no inverse.

Answer 4

Intuitively: $$1-\frac1{\frac1b+a}\cdot a=1-\frac{b}{1+ba}\cdot a=\frac{1+ba-ba}{1+ba}=\frac1{1+ba}.$$