# Unitary matrix norm

1. Feb 9, 2016

### pyroknife

1. The problem statement, all variables and given/known data
Prove $$||UA||_2 = ||AU||_2$$ where $U$ is a n-by-n unitary matrix and A is a n-by-m unitary matrix.

2. Relevant 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 $U^* = U^{-1}$

3. 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 in to the form of the 1st expression.

2. Feb 10, 2016

### Samy_A

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".

Last edited: Feb 10, 2016
3. Feb 10, 2016

### pyroknife

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.

4. Feb 10, 2016

### Samy_A

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?

5. Feb 10, 2016

### pyroknife

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

6. Feb 10, 2016

### Samy_A

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?

7. Feb 10, 2016

### pyroknife

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.

8. Feb 10, 2016

### Samy_A

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 $A^*A$.

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).

9. Feb 10, 2016

### pyroknife

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 $||UA||_2=||AU||_2$?

10. Feb 10, 2016

### Samy_A

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

11. Feb 10, 2016

### pyroknife

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

12. Feb 10, 2016

### Samy_A

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.

13. Feb 10, 2016

### pyroknife

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.