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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:


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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When $g=I$, the equality doesn't hold. – Davide Giraudo Jul 23 '12 at 11:42
It's somewhat jarring for me to see a matrix being denoted by a small letter... – J. M. Jul 23 '12 at 11:43
Are there any restrictions on the diagonal entries of $g$? E.g. if all diagonal entries vanish then $\det(I+g)=-\det(g-I)$ – Simon Markett Jul 23 '12 at 11:46
@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... – AndreasS Jul 23 '12 at 12:07
up vote 6 down vote accepted

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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Yes, thanks that helped a lot. There we also need that $|det(I-g)|^2=|det(g-I)|^2$, don't we? – AndreasS Jul 23 '12 at 12:19
Sure first use $\det(g-I)=(-1)^n\det(I-g),$ then change the variable $g\in U(n)\mapsto -g\in U(n).$ – Giuseppe Jul 23 '12 at 12:24

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