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April 4th, 2018, 03:42 AM  #1 
Newbie Joined: Mar 2018 From: Split, Croatia Posts: 13 Thanks: 0  Determinant of the conjugate matrix
I'm proving that detA=1 if A is unitary matrix. I got to the point where det(A*)det(A)=1. I know that det(A*)=det(A) conjugated (because determinant is polynomial with real coefficients), but I don't know how to precisely say that. Is there a simple way to prove that?

April 4th, 2018, 06:07 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
What you have written isn't true. For z a complex number, (z*)(z) is not equal to $\displaystyle z^2$. What is true is that $\displaystyle (z*)(z)= z^2$. You are missing the absolute values.

April 4th, 2018, 03:38 PM  #3 
Newbie Joined: Mar 2018 From: Split, Croatia Posts: 13 Thanks: 0  Thank you, it was an accidental mistake.. but can you answer my question?

April 4th, 2018, 05:04 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,981 Thanks: 790 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle det(A^*) det(A) = 1$ $\displaystyle det(A^{1}) det(A) = 1$ $\displaystyle \frac{1}{det(A)} \cdot det(A) = 1$ which is an identity. I don't recall the details of how to prove your theorem. I'll see if I can look it up tomorrow if someone doesn't help you out first. Dan  
April 4th, 2018, 06:04 PM  #5 
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics 
Unitary matrices preserve inner products (and thus norms). So every eigenvalue of a unitary matrix is on the complex unit circle. The determinant is the product of these eigenvalues so it must also lie on the complex unit circle.


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conjugate, determinant, matrix 
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