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Unitary Matrices Problem

  • Thread starter jumbogala
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  • #1
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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 = 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?
 

Answers and Replies

  • #2
Dick
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Look at <UX,UX>. You want to show that's equal to <X,X>. Use U^H=U^(-1).
 
  • #3
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I don't get how U^H = U^(-1) comes in here.

Maybe (UX)^H = (X^H)U^(-1), does that help?
 
  • #4
Dick
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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.
 
  • #5
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Sorry just for clarification, A^H and X are multiplied together in the above, correct? I can't find that in my textbook...
 
  • #6
Dick
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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.
 
  • #7
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Move the U on the right over to get [(UX)^H]U = X^H?
 
  • #8
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Wait, I think I got it!

<UX, UX > = <U^H UX, X> = <U^-1 UX, X> = <X, X>!
 
  • #9
Dick
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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>.
 
  • #10
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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?
 
  • #11
Dick
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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?
 
  • #12
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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
 
  • #13
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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:
  • #14
Dick
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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).
 
  • #15
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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?
 
  • #16
Dick
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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?
 
  • #17
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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.
 
  • #18
Dick
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Right. Not trying to prove UX=X.
 
  • #19
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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?
 
  • #20
Dick
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'*' 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.
 
  • #21
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Okay, I'll do that. Thank you very much for all your help!
 

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