Matrix Norms, and proving their characteristics

Question 1:

You know Ax is a vector right, and you know that the norm of a vector is greater than or equal to 0, with equality iff the vector is the zero-vector, right? So this is enough to show that ||A|| > 0. It remains to show that ||A|| = 0 iff A = 0.

Assume ||A|| = 0
Then max_{||x|| < 1}||Ax|| = 0
And then ||Ax|| = 0 for all x such that ||x|| < 1
From here, you need to prove that A = 0, and the above should be enough to make that easy.

On the other hand, assume A is 0 and prove that ||A|| = 0. That should be easy.

Question 2.

Your proof is not right. For one, you end up proving an inequality, and you're asked to prove an equality. But the overall approach is wrong. I assume that you are still working with the norm from the previous question, so:

||A|| = max_{||x|| < 1}||Ax||

You want to prove that ||A|| = max_{||x|| != 0}||Ax||/||x||, so you want to prove that:

max_{||x|| < 1}||Ax|| = max_{||x|| != 0}||Ax||/||x||

Let m = ||x||, and define v as x/m, so x = mv

max_{||x|| < 1}||Ax||
= max_{m < 1}m||Av||
= m x max_{m < 1}||Av||

max_{||x|| != 0}||Ax||/||x||
= max_{m != 0}m||Av||/m
= max_{m != 0}||Av||
= 1 x max_{m != 0}||Av||

All you have to do is show:

m x max_{m < 1}||Av|| = 1 x max_{m != 0}||Av||

Question 3.

Doesn't make any sense. On the right side of the equation, you seem to be "multiplying" two vectors because you have:

(x/||x||) x

Do you mean the dot product of (x/||x||) with x, or the cross product of those two vectors or, as I suspect, was this a typo?

 

Certainly you can see that (x/||x||)||x|| = x.

||x|| is the length of x, yes. If x is a 2-vector, then the set of x satisfying ||x|| < 1 is just the unit circle, so you are basically seeing how A transforms the unit circle, then finding the point in the image of the circle under the transformation A that is farthest from the origin.

 

If it helps, one of the more important properties of an operator norm is that it satisfies:

||Tx|| <= ||T|| ||x||

The definition of ||T|| is chosen so that ||T|| is large enough for this inequality to be valid, but no larger than necessary.

In fact, I've seen the definition written in three different ways (I'm not sure if I've seen the one you wrote or not): (click the image to see how to make them)

[tex]
||T|| := \sup_{x \neq 0} \frac{||Tx||}{||x||}
[/tex]

[tex]
||T|| := \sup_{||x|| \leq 1} ||Tx||
[/tex]

[tex]
||T|| := \sup_{||x|| = 1} ||Tx||
[/tex]

Exercise 1: prove that all three are equivalent.
Exercise 2: explain why these definitions capture the notion I gave above for why we choose the operator norm the way we do.