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

Question

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.

Answer 1

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

Answer 2

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 $I = U^{-1}U$:

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

Answer 3

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!