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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?
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April 4th, 2018, 05:07 AM   #2
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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.
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April 4th, 2018, 02:38 PM   #3
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Quote:
Originally Posted by Country Boy View Post
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?
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April 4th, 2018, 04:04 PM   #4
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Quote:
Originally Posted by Birgitta View Post
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
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April 4th, 2018, 05:04 PM   #5
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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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