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April 4th, 2018, 02:42 AM   #1
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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?
Attached Images 20180404_124200.jpg (98.0 KB, 12 views) April 4th, 2018, 05:07 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. Thanks from Birgitta April 4th, 2018, 02:38 PM   #3
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Quote:
 Originally Posted by Country Boy 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.
Thank you, it was an accidental mistake.. but can you answer my question? April 4th, 2018, 04:04 PM   #4
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Quote:
 Originally Posted by Birgitta 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?
Note: If A is unitary then we also know that $\displaystyle A^* = A^{-1}$. When you go from line 2 to line 3 then you need to have
$\displaystyle det(A^*) det(A) = 1$

$\displaystyle det(A^{-1}) det(A) = 1$

$\displaystyle \frac{1}{det(A)} \cdot det(A) = 1$

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, 05:04 PM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 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. Thanks from topsquark Tags conjugate, determinant, matrix Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post wortel Linear Algebra 1 January 6th, 2016 09:12 AM Favor Algebra 0 December 15th, 2015 08:25 AM jones123 Linear Algebra 3 April 24th, 2013 03:16 AM helloprajna Linear Algebra 2 December 8th, 2012 06:31 AM bull-roarer Algebra 2 May 23rd, 2009 02:44 PM

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