Eigenvalues of a unitary matrix

[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!

 

[Dick]

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

 

[kingwinner]

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

Multiply which of them together??

 

[quasar987]

The two equations.

 

[kingwinner]

The two equations.

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

 

[morphism]

=> ||X||=|lambda| ||U*X||

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

 

[kingwinner]

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

Why?

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

 

[HallsofIvy]

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.

 

[morphism]

Why?

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

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!

 

[Dick]

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

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