Understanding Vector Spaces: ||x||_inf and max |x_i| in R^n

squaremeplz
Messages
114
Reaction score
0

Homework Statement



Does ||x||_inf = max | x_i | for 1 <= i <= n define a norm on R^(n)

Homework Equations


The Attempt at a Solution



ok, I thought I understood vector spaces but this problem is confusing the heck out of me.

A norm is a function that assigns a positive and finite length to all vectors in a vector space.

so ||x||_inf = sqrt(x1^2 + x2^2 + ... x_inf)

max |x_i| depends on n in R^n

Can someone give me like a simple example? i.e. n = 2

Then ||x||_inf = max (|x1|, |x2|)?

The maximum distance between all vectors would be equal to the distance from 0 to the greatest vector. rather, infinity would be considered bounded by the max vector?

Any help is greatly appreciated.
 
Last edited:
Physics news on Phys.org
To determine if a function is a norm, you must check if is satisfies three axioms for all v,w in the vector space V:

||v|| >= 0 and equal to 0 iff v = 0.
|| v + w || <= || v || + || w ||
|| cv || = |c| || v ||,
where c is any scalar in the field the vector space is over.

Good Luck!
 
So the the question is asking whether max |x_i| for 1 <= i <= n satisfies these conditions?
 
Your definition of the norm is wrong. I think what you are writing (but x_{inf} should be squared and you should write \sum_{i=1}^\infty x_i^2- there is no &quot;x_{inf}&quot; member) is the &quot;Euclidean Norm&quot; (as you titled this thread) for l_n, the space of square summable sequences. The &quot;infimum&quot; norm on R<sup>n</sup> is MAX(x_1, x_2, \cdot\cdot\cdot, x_n).
 
Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top