# Unitary Matrices Problem

## Homework Statement

U is a unitary matrix. Show that
||UX|| = ||X|| for all X in the complex set.

Also show that |λ| = 1 for every eigenvalue λ of U.

## The Attempt at a Solution

I'm not sure where to start. So I looked up the definition of a unitary matrix. It satisfies one of these conditions:
U-1 = UH
The rows of U are an orthonormal set in the complex set
The columns of U are an orthonormal set in the complex set

Say X = [x1 x2 ... xn]

Now I know that ||X||2 = <X, X> = |x1|2 ... |xn|2

I'm not sure where to go from here. Can anyone help?

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
Look at <UX,UX>. You want to show that's equal to <X,X>. Use U^H=U^(-1).

I don't get how U^H = U^(-1) comes in here.

Maybe (UX)^H = (X^H)U^(-1), does that help?

Dick
Homework Helper
I don't get how U^H = U^(-1) comes in here.

Maybe (UX)^H = (X^H)U^(-1), does that help?
The hermitian adjoint satisfies <X,AY>=<A^H X,Y>. Move the U on the right over.

Sorry just for clarification, A^H and X are multiplied together in the above, correct? I can't find that in my textbook...

Dick
Homework Helper
Sorry just for clarification, A^H and X are multiplied together in the above, correct? I can't find that in my textbook...
Yes, A^H and X are multiplied together.

Move the U on the right over to get [(UX)^H]U = X^H?

Wait, I think I got it!

<UX, UX > = <U^H UX, X> = <U^-1 UX, X> = <X, X>!

Dick
Homework Helper
Move the U on the right over to get [(UX)^H]U = X^H?
No. No conjugate on X. <UX,UX>=<U^H UX,X>.

Okay, that makes sense. So now we have that's equal to <X, X>, so obviously their lengths are the same.

The only thing it says in my book though is that <VZ, W> = <Z, VW>... I wonder if I'm allowed to use the fact you gave me. Would it work using that instead?

Dick
Homework Helper
Okay, that makes sense. So now we have that's equal to <X, X>, so obviously their lengths are the same.

The only thing it says in my book though is that <VZ, W> = <Z, VW>... I wonder if I'm allowed to use the fact you gave me. Would it work using that instead?
<VZ,W>=<Z,VW> is only true if V is hermitian, i.e. V^H=V. That's not generally true for unitary matrices. Are you sure you are reading everything in the book, or are you just picking out formulas?

I missed the part where it said V has to be hermitian. I'm reading it but I guess I need to read more carefully...

I'm trying the next part now, showing that |λ| = 1.

Starting with UX = λX, I think I should be able to manipulate this to give λ = 1

I think I got the last part.
UX = λX
||UX|| = ||λX||
||UX|| = |λ| ||X||
||UX|| / ||X|| = |λ|

||X|| = ||UX|| so |λ| = 1

Going back to the first part of the question, the reason hermitian adjoint doesn't look familiar is because I haven't learned hilbert spaces yet. It's definitely not in there. Just out of interest, is there another way to do it?

If it's true that <UX,UX> = (UX)^H (UX), then I know how to do it.

Last edited:
Dick
Homework Helper
I think I got the last part.
UX = λX
||UX|| = ||λX||
||UX|| = |λ| ||X||
||UX|| / ||X|| = |λ|

||X|| = ||UX|| so |λ| = 1

Going back to the first part of the question, the reason hermitian adjoint doesn't look familiar is because I haven't learned hilbert spaces yet. It's definitely not in there. Just out of interest, is there another way to do it?

If it's true that <UX,UX> = (UX)^H (UX), then I know how to do it.
Sure, that shows |λ| = 1. The definition of <X,Y> is X^H Y. Since X^H is the conjugate transpose. So sure. <UX,UX> = (UX)^H (UX).

Okay, great. Then I think the first part can also be proved this way:
<UX, UX> = (UX)^H UX = X^H U^H U X = X^H U^(-1) U X = X^H X = <X,X>.

And necessarily if <UX, UX> = <X,X> their lengths are the same.

It's possible though for 2 vectors to have the same length but not be equal, isn't it?

Dick
Homework Helper
Okay, great. Then I think the first part can also be proved this way:
<UX, UX> = (UX)^H UX = X^H U^H U X = X^H U^(-1) U X = X^H X = <X,X>.

And necessarily if <UX, UX> = <X,X> their lengths are the same.

It's possible though for 2 vectors to have the same length but not be equal, isn't it?
You KNOW two vectors can have the same length and not be equal, right?

Whoops, I got confused. I forgot that <UX, UX> was directly related to the length of UX, and thought for a second that I was just proving that UX = X.

But yeah, [0 1] and [1 0] aren't equal but they have the same length.

Dick
Homework Helper
Right. Not trying to prove UX=X.

I just read somewhere that X*X = ||X||^2. I was thinking all along that X*X = ||X||, but I guess it doesn't affect my work.

* just means the same thing as H, right?

Dick