How Do Unitary Matrices Preserve Norms and Eigenvalue Magnitudes?

[jumbogala]

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.

Homework Equations

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 = U^H
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 ... x_n]

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

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

 

[Dick]

Look at <UX,UX>. You want to show that's equal to <X,X>. Use U^H=U^(-1).

 

[jumbogala]

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

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

 

[Dick]

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.

 

[jumbogala]

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

 

[Dick]

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.

 

[jumbogala]

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

 

[jumbogala]

Wait, I think I got it!

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

 

[Dick]

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>.

 

[jumbogala]

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]

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?

 

[jumbogala]

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

 

[jumbogala]

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.

 

[Dick]

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).

 

[jumbogala]

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]

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?

 

[jumbogala]

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]

Right. Not trying to prove UX=X.

 

[jumbogala]

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]

'*' means a lot of things. I'd stick with '^H'. And when you use it say 'I mean H is hermitian conjugate' (i.e. transpose and conjugate). The usual symbol is a superscript dagger. Always say what you think your symbols mean, especially if you aren't using TeX.

 

[jumbogala]

Okay, I'll do that. Thank you very much for all your help!