Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
When $g=I$, the equality doesn't hold. –  Davide Giraudo Jul 23 '12 at 11:42
1  
It's somewhat jarring for me to see a matrix being denoted by a small letter... –  J. M. Jul 23 '12 at 11:43
1  
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

1 Answer 1

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.

share|improve this answer
    
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 Tortorella Jul 23 '12 at 12:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.