# Homework Help: Norm of a vector problem.

1. Jul 20, 2011

### Oster

I'm trying to do a problem concerning converging sequences in normed linear spaces. Can anyone help me prove that if x=(x1,x2........,xn) is a vector in an n dimensional vector space then |xi| where i=1,2.....,n; is always less than or equal to ||x|| (norm of x). Maybe start out by writing x as a sum of n multiples of the basis vectors?

Last edited: Jul 20, 2011
2. Jul 20, 2011

### lanedance

its not quite true, should be less than or equal..

is this the usual norm or just a norm in general?

3. Jul 20, 2011

### Oster

yeah, i forgot to put in the "or equal to". It is any norm in general.

4. Jul 20, 2011

### lanedance

so what do you know about a norm that may help?

in particular, i would look at the triangle inequality

5. Jul 21, 2011

### lanedance

x1 = (x1,0,0,..) and
x = x1 + u
or maybe even better
x1 = x+(-u)

6. Jul 21, 2011

### Oster

uhhh
So I have ||x1|| = |x1| <= ||x|| + ||u||

7. Jul 22, 2011

### Oster

I still don't see how the triangle inequality implies this =/. I think I proved it using orthogonality and inner product properties in the finite dimensional case.

Assume the negation is true. There exists a non-zero vector x=(x1,x2.....xn) such that
|x1| > ||x|| writing out with an orthonormal basis, we get ||x1e1|| > ||x1e1.......+xnen||.
On squaring, we'd get x1^2 > x1^2 +...... + xn^2 whiich is a contradiction?

8. Jul 23, 2011

### lanedance

so this assumes the standard norm - is that ok?