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 Aisunitary!

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