Show that $A^{-1}-(A+C)^{-1}=A^{-1}(A^{-1}+C^{-1})^{-1}A^{-1}$

Question

I am asked to prove, for conformable non-singular matrices A and C, that $A^{-1}-(A+C)^{-1}=A^{-1}(A^{-1}+C^{-1})^{-1}A^{-1}$. However, I do not know of any general rules for inverse forms of a sum of matrices. Thus, I don't know how to expand out the terms in brackets using equivalence results. Without the summations, this seems like a straightforward problem: it seems I am missing something regarding the inverse of the sums. What might the first steps for moving past this be?

Answer 1

Since the right hand side of the desired identity can be written as $(A(A^{-1}+C^{-1})A)^{-1}$, you could try multiplying the left hand side by $A(A^{-1}+C^{-1})A = A + AC^{-1} A$ and show that the result equals the identity. Indeed you would get $$ (A^{-1} - (A + C)^{-1}) (A + AC^{-1} A) = I + \color{blue}{C^{-1} A} - \color{red}{(A+C)^{-1}A} - \color{green}{(A+C)^{-1} AC^{-1} A}\ . $$ The last term (green) can be rewritten as $(A+C)^{-1} (CA^{-1})^{-1} A = \color{orange}{(C + C A^{-1}C)^{-1}}A$. We can factor out a $A^{-1}C$ on the right from the (orange) terms in the parentheses, which allows us to write $$ (C + C A^{-1}C)^{-1} = [(A + C) (A^{-1}C)]^{-1} = C^{-1} A (A+C)^{-1}\ .$$ Plugging this back into the green term, we can now factor out $(A+C)^{-1} A $ from the right of the last two terms (red and green) in the sum, and get $$ (I+C^{-1}A)(A+C)^{-1} A = C^{-1}(C+A)(A+C)^{-1}A = C^{-1}A\ . $$ This will then cancel the second term (blue) of the sum. Thus the sum finally evaluates to $I$.

Thus to answer your "meta" question

What might the first steps for moving past this be?

I would say that

• when proving matrix identities of the form $A = B^{-1}$, try proving $AB = I$, and
• when proving identities involving inverse of sums, try "passing" factors on the outside from left to right or vice-versa, i.e. rewrite such factors as an inverse, multiply it into the sum, then factor out something on the other side.

Answer 2

This is a special case of the Woodbury matrix identity with both $U$ and $V$ being the appropriate identity matrix. You can find a proof of the general case there.

Answer 3

Step 1: In your equation $A^{-1}-(A+C)^{-1}=A^{-1}(A^{-1}+C^{-1})^{-1}A^{-1}$ multiply on the left by $A$ and on the right by $A + C$, to remove inverses from the left-hand side. You get the equivalent statement $$C = (A^{-1}+C^{-1})^{-1}A^{-1}(A + C)$$ Now multiply on the left by $A^{-1}+C^{-1}$ to clear the $(A^{-1}+C^{-1})^{-1}$ from the right-hand side. You get the equivalent statement $$(A^{-1} + C^{-1})(C) = A^{-1}(A + C)$$ This statement is immediate and we're done.