# Induced weighted matrix norm

1. Feb 20, 2016

### pyroknife

1. The problem statement, all variables and given/known data
The weighted vector norm is defined as
$||x||_W = ||Wx||$.
W is an invertible matrix.

The induced weighted matrix norm is induced by the above vector norm and is written as:
$||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W}$
A is a matrix.

Need to show $||A||_W = ||WAW^{-1}||$

2. Relevant equations

3. The attempt at a solution
$||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W} = sup_{x\neq 0} \frac{||WAx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}Wx||}{||Wx||}$

For an induced norm we know that:
$\frac{||WAW^{-1}Wx||}{||Wx||} \leq \frac{||WAW^{-1}||||Wx||}{||Wx||} = ||WAW^{-1}||$

Here is where I am lost. I already have gotten the expression into the form desired, but I do not know how to make the connection between $sup_{x\neq0}$ and the $\leq$. My thought is that we are taking the supremum of $\frac{||Ax||_W}{||x||_W}$, so it's maximum possible value is $||WAW^{-1}||$
and thus
$||A||_W=||WAW^{-1}||$
Is this the right logic?

2. Feb 20, 2016

### Staff: Mentor

The inequality holds for all $x \neq 0$ and therefore for the supremum, too.

3. Feb 21, 2016

### pyroknife

Yes I understand this, but I don't think this answers my question? Or maybe I do not actually understand?

4. Feb 21, 2016

### Staff: Mentor

Putting it together you already have:

$||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W} = sup_{x\neq 0} \frac{||WAx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}Wx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}Wx||}{||Wx||}$

$\leq sup_{x\neq 0} \frac{||WAW^{-1}||||Wx||}{||Wx||} = sup_{x\neq 0} \frac{||WAW^{-1}||||Wx||}{||Wx||} = sup_{x\neq 0} ||WAW^{-1}|| = ||WAW^{-1}||$

At least I cannot see why this shouldn't hold.

5. Feb 22, 2016

### pyroknife

Yes. I don't see anything wrong with those steps either, but I am not sure how to eventually get to the answer
$||A||_W = ||WAW^{-1}||$. Right now we have it in the form $||A||_W \leq ||WAW^{-1}||$

6. Feb 22, 2016

### Staff: Mentor

Right. I've forgotten the other direction. Give me some time to think about it.

7. Feb 22, 2016

### pyroknife

Thanks. I think it is some property of the "supremum" that I may be missing.

8. Feb 22, 2016

### Staff: Mentor

If we have $||A|| = sup_{x\neq 0} \frac{||Ax||}{||x||}$ which is true for the matrix norm induced by the vector norm, then it's easy (I think).

$||A||_W = sup_{x\neq 0} \frac{||Ax||_W}{||x||_W} = sup_{x\neq 0} \frac{||WAx||}{||Wx||} = sup_{y\neq 0} \frac{||WAW^{-1}y||}{||y||} = ||WAW^{-1}||$

9. Feb 22, 2016

### pyroknife

Hmm, I do not understand the step
$sup_{y\neq0}\frac{||WAW^{-1}y||}{||y||} = ||WAW^{-1}||$

Could you explain the equality?

10. Feb 22, 2016

### Staff: Mentor

The same as in the assumption that $||A|| = sup_{x\neq 0}{\frac{||Ax||}{||x||}}$ only with $||WAW^{-1}||$ instead of $||A||$ and $y$ instead of $x$.
Only point is whether this is right for the matrix norm given to you but usually it holds.