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Assuming that all matrix inverses involved below exist, show that $$(A-B)^{-1}=A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}A^{-1}.$$ In particular $$\left(I+A\right)^{-1}=I-\left(A^{-1}+I\right)^{-1}$$ and $$\left\vert\left(I+A\right)^{-1}+\left(I+A^{-1}\right)^{-1}\right\vert=1.$$

Here's what I tried to do but I can't simplify that underlined part at all:enter image description here

So I thought I should maybe try finding some sort of infinite geometric series in whose summation the $(AB^{-1} - I)$ term appears but that doesn't seem to help either and I'm not certain if that's mathematically justified either. What should I do next? Also, I've just started studying linear algebra so I would really appreciate an answer using only elementary properties of matrices.

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  • $\begingroup$ It may not be that straightforward. If I may, my advice is to first prove a formula for the inverse of X+Y whch may not depend on the inverse of the sum of inverses, apply the formula to compute $(X^{ -1}+Y^{-1})^{-1}$ and then tie those two equations together. $\endgroup$ Commented Dec 31, 2023 at 6:41
  • $\begingroup$ I expect that someone will point out that you should use LaTeX for the question and not an image. This makes the question searchable. $\endgroup$
    – Blitzer
    Commented Dec 31, 2023 at 6:55
  • $\begingroup$ Yeah sorry about that, I'm new to all this and I don't know how to use latex at all $\endgroup$ Commented Dec 31, 2023 at 7:19

1 Answer 1

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It is easy to note that

$(A-B)\big(A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}A^{-1}\big) \\= I + (B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1}(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1} ~(\text{as}~ AA^{-1}=I) \\= I + BB^{-1}(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1}(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1} ~(\text{as}~ BB^{-1}=I) \\= I + B(B^{-1} - A^{-1})(B^{-1} - A^{-1})^{-1}A^{-1} - BA^{-1} \\= I + BA^{-1} - BA^{-1} = I$

That implies $(A-B)^{-1} = \big(A^{-1} + A^{-1}(B^{-1} - A^{-1})^{-1}A^{-1}\big)$

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  • $\begingroup$ thank you for such a clear answer! $\endgroup$ Commented Dec 31, 2023 at 7:20
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    $\begingroup$ You are welcome :) $\endgroup$
    – Afntu
    Commented Dec 31, 2023 at 9:11

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