# Does the identity $\det(I+g^{-1})\det(I+g)=|\det(g-I)|^2$ hold for $g \in U(n)$?

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.

• When $g=I$, the equality doesn't hold. Jul 23, 2012 at 11:42
• It's somewhat jarring for me to see a matrix being denoted by a small letter... Jul 23, 2012 at 11:43
• Are there any restrictions on the diagonal entries of $g$? E.g. if all diagonal entries vanish then $\det(I+g)=-\det(g-I)$ Jul 23, 2012 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... Jul 23, 2012 at 12:07

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 $|\det(I+g)|^{2k}.$
• Yes, thanks that helped a lot. There we also need that $|det(I-g)|^2=|det(g-I)|^2$, don't we? Jul 23, 2012 at 12:19
• Sure first use $\det(g-I)=(-1)^n\det(I-g),$ then change the variable $g\in U(n)\mapsto -g\in U(n).$