Are the Eigenvalues of a Unitary Operator of the Form e^i(a)?

[jnazor]

Homework Statement

A unitary operator U has the property
U(U+)=(U+)U=I [where U+ is U dagger and I is the identity operator]

Prove that the eigenvalues of a unitary operator are of the form e^i(a) with a being real.

NB: I haven't been taught dirac notation yet. Is there a way i can do this without it?

Homework Equations

U(U+)=(U+)U=I [where U+ is U dagger and I is the identity operator]

The Attempt at a Solution

Assume eigenvalues exist
U(a)=x(a) => (U+)U(a)=(U+)x(a) => (a)=(U+)x(a)??

 

[mjsd]

note that you don't need to understand Dirac notation, all you need to know is some basic linear algebra in finite dimensional space. hint: "of the form $$e^{i\theta}$$" means that magnitude of complex e-vals are 1

 

[dextercioby]

HINT: U unitary means U isometry. Assume the spectral equation

$$U\psi =a\psi$$ (1)

has solutions in a Hilbert space $\mathcal{H}$.

Then use (1), the assumption regarding the space of solutions and the isometry condition to get the desired result.

 

[jnazor]

Sorry I've never heard of isometry or the name spectral equation. I just know it as the eigenvalue equation.

 

[Dick]

Isometry means <x,y>=<Ux,Uy>. Why is this true for U unitary? Once you believe it's true set y=x and x to be an eigenvector of U. What do you conclude?