Unitary Matrices and Eigenvalues: Exploring Properties and Applications

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In summary, the conversation discusses how to show that a unitary matrix U satisfies ||UX|| = ||X|| for all X in the complex set and that |λ| = 1 for every eigenvalue λ of U. The conversation includes various attempts at a solution, including using the definition of a unitary matrix and manipulating equations, as well as discussing the use of the hermitian adjoint in proving the first part. Through these attempts, it is concluded that the first part can also be proved using the definition of <X,Y>, and that the second part can be shown by manipulating the equation UX = λX. Finally, it is acknowledged that two vectors can have the same length but not be equal, and that clarification is
  • #1
jumbogala
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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?
 
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  • #2
Look at <UX,UX>. You want to show that's equal to <X,X>. Use U^H=U^(-1).
 
  • #3
I don't get how U^H = U^(-1) comes in here.

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

<UX, UX > = <U^H UX, X> = <U^-1 UX, X> = <X, X>!
 
  • #9
jumbogala said:
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
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
jumbogala said:
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
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
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
jumbogala said:
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
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
jumbogala said:
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
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
Right. Not trying to prove UX=X.
 
  • #19
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
'*' 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
Okay, I'll do that. Thank you very much for all your help!
 

1. What is a unitary matrix?

A unitary matrix is a square matrix whose conjugate transpose is equal to its inverse. In other words, a unitary matrix is a square matrix that preserves the length of its vectors and the angles between them.

2. Why are unitary matrices important in mathematics?

Unitary matrices have many important applications in mathematics, particularly in linear algebra and complex analysis. They are used to represent rotations and reflections in Euclidean spaces, and they also play a crucial role in quantum mechanics.

3. How are unitary matrices related to orthogonal matrices?

Unitary matrices are the complex analogue of orthogonal matrices. Both types of matrices have the property that their transpose is equal to their inverse. However, orthogonal matrices are only defined for real numbers, while unitary matrices are defined for complex numbers.

4. Can unitary matrices be used to solve systems of linear equations?

Yes, unitary matrices can be used to solve systems of linear equations. In fact, they are often used in conjunction with other matrix operations to efficiently solve systems of equations.

5. Are all unitary matrices invertible?

Yes, all unitary matrices are invertible because their conjugate transpose is equal to their inverse. This means that there exists a matrix that, when multiplied by the given unitary matrix, will result in the identity matrix.

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