# Eigenvalues of a unitary matrix

1. Dec 11, 2007

### kingwinner

Q: Prove htat if a matrix U is unitary, then all eigenvalues of U have absolute value 1.

My try:
Suppose U*=U^-1 (or U*U=I)

Let UX=(lambda)X, X nonzero
=> U*UX=(lambda) U*X
=> X=(lambda) U*X
=> ||X||=|lambda| ||U*X||
=> |lambda| = ||X|| / ||(U^-1)X||

And now I am really stuck and hopeless, what can I do?

Thanks for helping!

Last edited: Dec 11, 2007
2. Dec 11, 2007

### Dick

If Ux=(lambda)x then (x*)(U*)=(lambda*)(x*). Multiply those together.

3. Dec 12, 2007

### kingwinner

Multiply which of them together??

4. Dec 12, 2007

### quasar987

The two equations.

5. Dec 12, 2007

### kingwinner

OK, so
Ux (x*)(U*) = (lambda)x (lambda*)(x*)
But how does this prove the statement?

6. Dec 12, 2007

### morphism

If U is unitary, then ||Ux||=||x||.

7. Dec 12, 2007

### kingwinner

Why?

Also, I know nothing about isomertry. Is there any way to prove the statement without this concept? Can somone please show me how?

8. Dec 12, 2007

### HallsofIvy

Staff Emeritus
Your initial definition of "unitary" was U*U= I, right? If $\lambda$ is an eigenvalue of U, with corresponding eigenvector v, then $Uv= \lambda v$. But then $U*U(v)= \lambda U*(v)$ and since U*U= I, U*U(v)= v. That is, for any eigenvalue $\lambda$ of U and corresponding eigevector v, $\lambda U*(v)= v$. That means that v is also an eigenvector of U* with eigenvalue $1/\lambda$.

9. Dec 12, 2007

### morphism

On second thought, maybe you should stick to what the other guys are suggesting. (For me, unitary = special kind of operator on an inner product space. But if you're working strictly with matrices, this might not be helpful. [For the sake of completeness: ||Ux||^2 = <Ux,Ux> = <x,U*Ux> = <x,x> = ||x||^2.])

Just thought I'd also mention lambda is never going to be zero (because U is invertible), so we can safely say things like 1/lambda!

Last edited: Dec 12, 2007
10. Dec 12, 2007

### Dick

Multiply them in the OTHER ORDER, so you get a (U*)(U).