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I am currently working on a problem and I am stuck with the following issue.

For $A \in GL(n)$ and $B \in U(n)$ I am hoping that it is true that

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

My question is whether this is indeed the case and if so what I need to look into to understand why. ( I just assume for the moment that $B - A$ has an inverse )

PS: Just to give some context I am stuck with this issue because I am playing around with the kernel that I have computed for the operator

$$ D := i \;I\frac{d}{dx} + B(x) : C^\infty_T ([0,1],\mathbb{C^m}) \to C^\infty_T ([0,1],\mathbb{C^m})$$ with boundary condition $f(1) = T f(0)$ for $T \in U(n)$ and $B$ Hermitian.

I am currently trying to show that $L_T \;f(1) = T L_T \; f(0)$ where $L_T$ is the integral operator given by the kernel that I have computed for $D$. In order for things to work out I have to show that

$$ p(1)(T-p(1))^{-1}T = T(T-p(1))^{-1}p(1)$$ where $p$ is assumed to conjugate $D$ to $-ip^{-1}Dp=D^p = \frac{d}{dx}I$

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How do you know that $B-A$ has an inverse? – Fabian Dec 31 '12 at 13:43
good point, thanks. It's an assumption for the moment. – Beltrame Dec 31 '12 at 13:51
thanks guys, seems the context was not necessary ... :) – Beltrame Dec 31 '12 at 14:35
up vote 2 down vote accepted

Hint: try to show that both expressions are in fact given by $$(A^{-1}-B^{-1})^{-1};$$ we have to assume that $A$, $B$, $B-A$ are invertible (no unitary requirement for $B$ needed).

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+1 This is not a competition for speed. Happy new year! :-D – user1551 Dec 31 '12 at 14:27

$A (B-A)^{-1} B=[B^{-1}(B-A)A^{-1}]^{-1}=(A^{-1}-B^{-1})^{-1}$. Similarly $B (B-A)^{-1} A=[A^{-1}(B-A)B^{-1}]^{-1}=(A^{-1}-B^{-1})^{-1}$. So they are equal as long as $A,B$ and $B-A$ are invertible.

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+1 you were a bit faster than me... – Fabian Dec 31 '12 at 14:24

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