Proving Unitary Matrix Norm: $$||UA||_2 = ||AU||_2$$

[pyroknife]

Homework Statement

Prove $$||UA||_2 = ||AU||_2$$ where $U$ is a n-by-n unitary matrix and A is a n-by-m unitary matrix.

Homework Equations

For any matrix A, $||A||_2 = \rho(A^*A)^.5$, $\rho$ is the spectral radius (maximum eigenvalue)
where $A^*$ presents the complex conjugate of A.

U is unitary, which means $U^* = U^{-1}$

The Attempt at a Solution

$||UA||_2 = \rho(A^* U^* UA)^.5 = \rho(A^* A)^.5$
$||AU||_2 = \rho(U^* A^* AU)^.5$
This is really as far as I got. I know that I must prove how to equate these 2 expressions. But the second expression is giving me some trouble, and nothing jumps out as me as to how to get this into the form of the 1st expression.

 

[Samy_A]

Homework Statement

Prove $$||UA||_2 = ||AU||_2$$ where $U$ is a n-by-n unitary matrix and A is a n-by-m unitary matrix.

Homework Equations

For any matrix A, $||A||_2 = \rho(A^*A)^.5$, $\rho$ is the spectral radius (maximum eigenvalue)
where $A^*$ presents the complex conjugate of A.

U is unitary, which means $U^* = U^{-1}$

The Attempt at a Solution

$||UA||_2 = \rho(A^* U^* UA)^.5 = \rho(A^* A)^.5$
$||AU||_2 = \rho(U^* A^* AU)^.5$
This is really as far as I got. I know that I must prove how to equate these 2 expressions. But the second expression is giving me some trouble, and nothing jumps out as me as to how to get this into the form of the 1st expression.

Try to show that an eigenvalue of $A^* A$ is also an eigenvalue of $U^* A^* AU$, and likewise that an eigenvalue of $U^* A^* AU$ is also an eigenvalue of $A^* A$.

EDIT: in your problem statement, it says "A is a n-by-m unitary matrix". I assume that was meant to be "A is a n-by-n matrix".

 

[pyroknife]

Try to show that an eigenvalue of $A^* A$ is also an eigenvalue of $U^* A^* AU$, and likewise that an eigenvalue of $U^* A^* AU$ is also an eigenvalue of $A^* A$.

EDIT: in your problem statement, it says "A is a n-by-m unitary matrix". I assume that was meant to be "A is a n-by-n matrix".

Thanks, this is what I am trying to do, but I am lost on how to approach this. There must be some special property unitary matrices that I am missing that could greatly help. Any hints?

Regarding your edit: No, my OP is correct. A is an n-by-m matrix where n doesn't necessarily have to equal m. I was confused about this at first because we are dealing with eigenvalues which only works with square matrices, but the 2-norm of a matrix is the max eigenvalue of the matrix $A^*A$ which is always square.

 

[Samy_A]

Regarding your edit: No, my OP is correct. A is an n-by-m matrix where n doesn't necessarily have to equal m. I was confused about this at first because we are dealing with eigenvalues which only works with square matrices, but the 2-norm of a matrix is the max eigenvalue of the matrix $A^*A$ which is always square.

I don't get this. If A is a n-by-m matrix with m different from n, how do you define $AU$, the multiplication of a n-by-m matrix by a n-by-n matrix?

 

[pyroknife]

I don't get this. If A is a n-by-m matrix with m different from n, how do you define $AU$, the multiplication of a n-by-m matrix by a n-by-n matrix?

Hmmm, I completely missed that part. I agree. This is a strange. A has to be square then.

 

[Samy_A]

Hmmm, I completely missed that part. I agree. This is a strange. A has to be square then.

Ok.

So just start with an eigenvalue $\lambda$ of $U^* A^* AU$.
For some $x \neq \vec 0$, $U^* A^* AUx=\lambda x$.
Now Apply $U$ to both sides, and remember that $U$ is unitary, meaning $UU^*=I$.

Conversely, say $\mu$ is an eigenvalue of $A^*A$, meaning $A^*Ay=\mu y$ for some $y \neq \vec 0$. Since $U$ is invertible, there is a $z$ such that $y=Uz$. That yields $A^*AUz=\mu Uz$. What happens when you apply $U^*$ to both sides?

 

[pyroknife]

Ok.

So just start with an eigenvalue $\lambda$ of $U^* A^* AU$.
For some $x \neq \vec 0$, $U^* A^* AUx=\lambda x$.
Now Apply $U$ to both sides, and remember that $U$ is unitary, meaning $UU^*=I$.

Conversely, say $\mu$ is an eigenvalue of $A^*A$, meaning $A^*Ay=\mu y$ for some $y \neq \vec 0$. Since $U$ is invertible, there is a $z$ such that $y=Uz$. That yields $A^*AUz=\mu Uz$. What happens when you apply $U^*$ to both sides?

Thanks! So when you apply $U^*$ to both sides, you get the expression:
$U^*A^*AUz = \mu z$
and we know $z = U^{-1}y$, plugging this back in, we obtain:
$U^*A^*AUU^{-1}y = \mu U^{-1}y$
=> $U^*A^*Ay = \mu U^{-1}y$
Since U is a unitary matrix, that means $U^* = U^{-1}$, so the above expression becomes (after multiplying U to both sides):
$A^*Ay = \mu y$

This is not sufficient right? What I see is I have shown that $A^*Ay=\mu y$ and $U^*A^*AUz = \mu z$ have the same eigenvalue $\mu$ but corresponding to different eigenvectors.

 

[Samy_A]

Thanks! So when you apply $U^*$ to both sides, you get the expression:
$U^*A^*AUz = \mu z$

You can stop here. You have shown that $\mu$ is an eigenvalue of $U^*A^*AU$ (with eigenvector $z$, but we don't care much for eigenvectors in this exercise, see below). That goes for any eigenvalue of $A^*A$.

This is not sufficient right? What I see is I have shown that $A^*Ay=\mu y$ and $U^*A^*AUz = \mu z$ have the same eigenvalue $\mu$ but corresponding to different eigenvectors.

The spectral radius depends on the eigenvalues, not the eigenvectors.
If you show that $U^*A^*AU$ and $A^*A$ have the same eigenvalues, it follows that their spectral radius is the same too. That then proves that $||UA||_2=||AU||_2$ (see your first post).

 

[pyroknife]

You can stop here. You have shown that $\mu$ is an eigenvalue of $U^*A^*AU$ (with eigenvector z, but we don't care much for eigenvectors in this exercise, see below). That goes for any eigenvalue of $A^*A$.

Now do the same in the reverse order, with some $\lambda$ as eigenvalue of $U^*A^*AU$. Show that it is an eigenvalue of $A^*A$.
The spectral radius depends on the eigenvalues, not the eigenvectors.
If you show that $U^*A^*AU$ and $A^*A$ have the same eigenvalues, it follows that their spectral radius is the same too. That then proves that $||UA||_2=||AU||_2$ (see your first post).

Great thanks.
Why do we need to prove it in the reverse order?
Typically for if and only if conditions I have to prove both in the forward and reverse direction, but this seems like just an equality proof. Shouldn't showing that $\mu$ is an eigenvalue of both $U^*A^*AU$ and $A^*A$ already have $||UA||_2=||AU||_2$?

 

[Samy_A]

Great thanks.
Why do we need to prove it in the reverse order?
Typically for if and only if conditions I have to prove both in the forward and reverse direction, but this seems like just an equality proof. Shouldn't showing that $\mu$ is an eigenvalue of both $U^*A^*AU$ and $A^*A$ already have $||UA||_2=||AU||_2$?

Yes, I see now how you did it. It is somewhat different from what I had in mind, but equally correct.

 

[pyroknife]

Yes, I see now how you did it. It is somewhat different from what I had in mind, but equally correct.

Awesome, what were you thinking out of curiosity? I was trying proving the reverse way earlier, but couldn't figure it out.

 

[Samy_A]

Awesome, what were you thinking out of curiosity? I was trying proving the reverse way earlier, but couldn't figure it out.

The reverse part I had in mind was:
Let $\lambda$ be an eigenvalue of $U^* A^* AU$, then $U^* A^* AUx=\lambda x$.
Applying $U$ to both sides gives $UU^* A^* AUx=\lambda Ux$. Since $U$ is unitary, this becomes $A^* AUx=\lambda Ux$, showing that $\lambda$ is an eigenvalue of $A^* A$ (with eigenvector $Ux$).
But this is essentially the same as what you did.

 

[pyroknife]

The reverse part I had in mind was:
Let $\lambda$ be an eigenvalue of $U^* A^* AU$, then $U^* A^* AUx=\lambda x$.
Applying $U$ to both sides gives $UU^* A^* AUx=\lambda Ux$. Since $U$ is unitary, this becomes $A^* AUx=\lambda Ux$, showing that $\lambda$ is an eigenvalue of $A^* A$ (with eigenvector $Ux$).
But this is essentially the same as what you did.

OH WOW. I actually got to the expression $A^* AUx=\lambda Ux$, but I was stuck on the $Ux$. I was under the impression that the eigenvector x from the original expression should be maintained, but as we discussed, that is not the case since we really only care about the eigenvalues. And I also failed to recognize that $Ux$ is an eigen-vector itself corresponding to $A^*A$.

Thanks for all the help.