# Is this a valid step?||Ax|| + ||Bx|| <= ||A||*||x|| + ||B||*||x||

1. Apr 22, 2007

### EvLer

Is this a valid step?

||Ax|| + ||Bx|| <= ||A||*||x|| + ||B||*||x|| ??

where A and B are actually matrices, x is a vector but actually it can be reduced to just numbers and a variable, respectively. And ||x|| = sqrt (x*x)

thanks as always :)

2. Apr 22, 2007

### AKG

It suffices to know that if M is any matrix and v is any vector (such that Mv is defined), then $||Mv|| \leq ||M||\, ||v||$. What's the definition of $||M||$?

3. Apr 22, 2007

### EvLer

I don't know.... what I actually have to prove is that ||A+B|| <= ||A|| + ||B|| follows from a norm of a matrix A:
||A|| = max (||Ax||/||x||)

4. Apr 22, 2007

### ZioX

||A+B||=max(|(A+B)x|/|x|=max(|Ax+Bx|)/|x| and Ax and Bx are just vectors....so you can just use the triangle inequalities.

Also, do note, your defintion for norm is wrong. You have to some constraint on x. The most common is |x|<=1.

Alos, I don't know if you realize this or not, but you're suppose to be taking the vector norm of Ax divided by the vector norm of x.

5. Apr 23, 2007

### AKG

Well what don't you get? You want to prove:

$$||Ax|| + ||Bx|| \leq ||A||\, ||x|| + ||B||\, ||x||$$

to do this, it suffices to prove:

$$||Ax|| \leq ||A||\, ||x||$$

and

$$||Bx|| \leq ||B||\, ||x||$$

right? But there's nothing special about A and B, all you really need to show is that for any matrix M and any vector v that:

$$||Mv|| \leq ||M||\, ||v||$$

right? Well, if v is 0, then the inequality holds (and in fact, it's an equality, 0=0). If v is non-zero, then the above is equivalent to:

$$||Mv||/||v|| \leq ||M||$$

But the very definition of the norm of a matrix is:

$$\sup _{v \neq 0} ||Mv||/||v||$$

(which is pretty much the definition you gave, except you used "max" instead of "sup" which isn't a big deal, and you forgot to rule out x=0, since in that case $||Ax||/||x||$ is undefined). So when v is non-zero,

$$||M|| \geq ||Mv||/||v||$$

by the very definition of the matrix norm.