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$S$ is real symmetric. $T$ is real skew-symmetric. I have shown that $T\pm iS$ is skew-Hermitian. I am further asked to show that $U = (I+T+iS)(I-T-iS)^{-1}$ is unitary.

Denoting by $^\dagger$ the conjugate transpose, I have that
$U^\dagger = [(I+T+iS)(I-T-iS)^{-1}]^\dagger$
$= ((I-T-iS)^\dagger)^{-1}(I+T+iS)^\dagger$
$= (I+T+iS)^{-1}(I-T-iS)$

But then the products $UU^\dagger$ and $U^\dagger U$ have the factors in the wrong order to cancel out to make identity. I must be missing something obvious here; I'd appreciate a steer!

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For any $A$, $(I+A)$ commutes with $(I-A)$, and hence with $(I-A)^{-1}$ if it exists. – Chris Eagle Jan 5 '11 at 19:25
Thanks Chris, that makes it easy. I don't think I've seen that simple (but useful-looking) result until now! I have proved it using index notation; is there a nicer way? – user5426 Jan 5 '11 at 19:50
If $A$ commutes with $B$, then $AB=BA$. If further $B$ is invertible, then $B^{-1}(AB)B^{-1}=B^{-1}(BA)B^{-1}$, and hence $B^{-1}A=AB^{-1}$. – Chris Eagle Jan 5 '11 at 19:58
Sorted, thanks very much indeed. – user5426 Jan 6 '11 at 1:09

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