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In a paper (corollary 1, p.14) the following identity is used:

Let g be a unitary matrix. Then:

$\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2 \text{ for }g \in U(n)$

Now my question is why this holds

I calculated:

$\det(I+g^{-1})\det(I+g)=\overline{\det(I+g^t)}\det(I+g)=\overline{\det(I+g)}\det(I+g)=|\det(I+g)|^2$

Where the second equality holds as $I$ has only entries in the diagonal ($I$ is of course the unit matrix). But this is not the same as on the right side.

(I also thought that maybe there was a typo on the left side where should be minus-signs. However in the paper itself it is needed that there are plus-signs.)

Thanks for any hints.

Edit: This equality was in the scope of an integral: $\int_{U(n)}\prod_{l=1}^{k}det(I+g^{-1})\prod_{l=1}^{k}\det(I+g)dg=\int_{U(n)}|\det(g-I)|^{2k}dg$

With a change of variable it was solved with my calculation done above. See Giuseppe's answer.

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    @DavideGiraudo Yes, you're completely right. I didn't see it... So I assume there is indeed a sign issue here - I have to look again at this corollary...2012-07-23

1 Answers 1

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Dear AndreasS I have given a look at the paper.

I think that the calculation $\det(I+g^{-1})\det(I+g)=|\det(I+g)|^2 \text{ for }g \in U(n)$ is correct.
But in order to obtain the result stated in Corollary 1, you just need the change of variable $g\mapsto -g$ in the integral over $U(n)$, so that $\int_{U(n)}|\det(g-I)|^{2k}dg=\int_{U(n)}|\det(I+g)|^{2k}dg.$ Then in the paper you find how factorize $|\det(I+g)|^{2k}.$

I hope that it helps.

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    Sure first use $\det(g-I)=(-1)^n\det(I-g),$ then change the variable $g\in U(n)\mapsto -g\in U(n).$2012-07-23