Question

Proof, if A is a unitary matrix, then det(A)=1

Answer 1

I tried to prove it to \[A\epsilon \mathbb{C}^{2x2}\] but I didn't know how to do generalize the idea to an nxn case.

Answer 2

Are you sure it isn't |det(A)| = 1?

Answer 3

Well, anyway, I don't think it's true unless |det(A)| = 1. Here's the proof for what I stated though:

1 = det(I) = det(AA*) = det(A)det(A*) = det(A)det(A)* = |det(A)|^2

So, |det(A)| = 1

Answer 4

If det(A)=1 then det(A)=1... it's trivial.
Anyway thank you, I found the same proof using some facts about determinants I found in wikipedia.

Answer 5

|det(A)| = 1 => det(A) = 1 or -1. It isn't trivial. The absolute value is required.

Answer 6

Sorry I misread.

Answer 7

I misread 2 times, one was when I read "I don't think it's true unless |det(A)| = 1"
And then when I read the exercise in my homework.
It was |det(A)|, not det(A) :p

Answer 8

Ah. Good. Just checking. :)

Answer 9

I'm closing this.