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

If $$A$$ and $$B$$ are square matrices, $$A+B$$ is invertible, and $$(A+B)^{-1}=A^{-1} + X$$, show that the matrix $$X$$ can be written as: $$X=-(I+A^{-1}B)^{-1}A^{-1}BA^{-1}$$.

I've tried to show that:

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

And I arrived in:

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

I don't know what to do now and I didn't find any property that helps me.

• Hi and welcome to the site! It is advisable that you take a tour to see what we are about. Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. Here's a quick guide (if nothing else, read up the part on "avoiding no-clue questions").
– 5xum
Commented Mar 16, 2023 at 12:24
• Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. Even if the question is closed, you can still edit it, and we will vote to reopen it.
– 5xum
Commented Mar 16, 2023 at 12:24
• "inversible"? maybe you meant invertible?
– D S
Commented Mar 16, 2023 at 12:41
• I think you need to show that $(A+B)[A^{-1}-(I+A^{-1}B)^{-1}A^{-1}BA^{-1}]=I$, not the identity you wrote. Commented Mar 16, 2023 at 12:47
• shouldn't you be proving that $(A+B)(A^{-1}+X) = I$?
– D S
Commented Mar 16, 2023 at 12:49

You have $$X=(A+B)^{-1}-A^{-1}$$.

Step 1, take out $$A$$ from the first summand: $$X= [A(I+A^{-1}B)]^{-1}-A^{-1}=(I+A^{-1}B)^{-1}A^{-1}-A^{-1}=[(I+A^{-1}B)^{-1}-I]A^{-1}$$

Step 2, take out $$(I+A^{-1}B)^{-1}$$: $$X=(I+A^{-1}B)^{-1}[I-(I+A^{-1}B)]A^{-1}=(I+A^{-1}B)^{-1}(-A^{-1}B))A^{-1}$$

First rearrange the equation:

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

Next, use the the properties $$(UV)^{-1}=V^{-1}U^{-1}$$ for invertible matrices $$U,V$$ and the distributive property:

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

Finally, use the fact that for an invertible matrix $$U$$ we have $$I = U^{-1}U$$:

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

Going blindly in: we multiply the candidate from the right: $$[A^{-1}-(I+A^{-1}B)^{-1}A^{-1}BA^{-1}](A+B)$$ $$=I + A^{-1}B-(I+A^{-1}B)^{-1}A^{-1}B-(I+A^{-1}B)^{-1}A^{-1}BA^{-1}B$$ Denote $$X=A^{-1}B$$ to rewrite $$=I + X-(I+X)^{-1}X-(I+X)^{-1}X^2$$ $$=I + (I-(I+X)^{-1}-(I+X)^{-1}X)X$$

$$=I + (I-(I+X)^{-1}(I+X))X = I+(I-I)X = I$$ Done!