# Thread: Proof of Eigenvalues of unitary matrix lie on unit circle

1. ## Proof of Eigenvalues of unitary matrix lie on unit circle

I'm really frustrated because I first thought this would be simple. Here's what I have so far.

Designate the matrix A, the eigenvalues x (since I can't TeX). Then essentially we wish to show the ||x|| = 1, meaning more specifically that x* (x-conjugate) = 1 for all eigenvalues x.

Getting there is tricky for me though. Obviously since A is unitary, the product of A with its conjugate transpose is commutatively equal to the identity. Basic algebra yields that the Inverse of A is equal to its conjugate transpose. But I have no rules about the products of matrices, esp. complex matrices to use.

I'm looking at the spectral theorem for normal operators, as the subject is in the same section, but I'm not seeing any helpful tie ins. Any help would be great.

2. This is very difficult to understand. You start by simply asserting that you want to prove that the eigenvalues of matrix A have length 1. But you haven't said what A is!

Later you say "Obviously since A is unitary"- but you still haven't said what A is, so it is certainly NOT obvious that A is unitary!

Okay, I now see that your title specifically said that you are trying to prove that the eigenvalues of any unitary matrix lie on the unit circle. (In general, it is a bad idea not to state the question in full in the body of the post.)

Let v be an eigenvector of A with eigenvalue $\lambda$: [tex]Av=\lambda v[tex] so that $A*Av= A*(\lambda v)= \lambda (A*v)= v$. Now, if you know that an eigenvector of A, with eigenvalue $\lambda$, is also an eigenvector of $A*$, with eigenvalue $\overline{\lambda}$, it will be easy to finish. Do you have that theorem or can you prove it?

3. Take a look at this thread.

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### eigenvactor lying on unit circle

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