Difference between revisions of "Norm"
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<math>\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}</math> | <math>\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}</math> | ||
+ | ====Proof that it is a norm==== | ||
{{Todo|proof}} | {{Todo|proof}} | ||
+ | =====Part 4 - Triangle inequality===== | ||
+ | Let <math>x,y\in\mathbb{R}^n</math> | ||
+ | |||
+ | <math>\|x+y\|_2^2=\sum^n_{i=1}(x_i+y_i)^2</math> | ||
+ | <math>=\sum^n_{i=1}x_i^2+2\sum^n_{i=1}x_iy_i+\sum^n_{i=1}y_i^2</math> | ||
+ | <math>\le\sum^n_{i=1}x_i^2+2\sqrt{\sum^n_{i=1}x_i^2}\sqrt{\sum^n_{i=1}y_i^2}+\sum^n_{i=1}y_i^2</math> using the [[Cauchy-Schwarz inequality]] | ||
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+ | <math>=\left(\sqrt{\sum^n_{i=1}x_i^2}+\sqrt{\sum^n_{i=1}y_i^2}\right)^2</math> | ||
+ | <math>=\left(\|x\|_2+\|y\|_2\right)^2</math> | ||
+ | |||
+ | Thus we see: <math>\|x+y\|_2^2\le\left(\|x\|_2+\|y\|_2\right)^2</math>, as norms are always <math>\ge 0</math> we see: | ||
+ | |||
+ | <math>\|x+y\|_2\le\|x\|_2+\|y\|_2</math> - as required. | ||
+ | |||
{{Definition|Linear Algebra}} | {{Definition|Linear Algebra}} |
Revision as of 17:15, 7 March 2015
Contents
[hide]Definition
A norm on a vector space (V,F) is a function ∥⋅∥:V→R such that:
- ∀x∈V ∥x∥≥0
- ∥x∥=0⟺x=0
- ∀λ∈F,x∈V ∥λx∥=|λ|∥x∥ where |⋅| denotes absolute value
- ∀x,y∈V ∥x+y∥≤∥x∥+∥y∥ - a form of the triangle inequality
Often parts 1 and 2 are combined into the statement
- ∥x∥≥0 and ∥x∥=0⟺x=0 so only 3 requirements will be stated.
I don't like this
Examples
The Euclidean Norm
The Euclidean norm is denoted ∥⋅∥2
Here for x∈Rn we have:
∥x∥2=√n∑i=1x2i
Proof that it is a norm
TODO: proof
Part 4 - Triangle inequality
Let x,y∈Rn
∥x+y∥22=n∑i=1(xi+yi)2 =n∑i=1x2i+2n∑i=1xiyi+n∑i=1y2i ≤n∑i=1x2i+2√n∑i=1x2i√n∑i=1y2i+n∑i=1y2i using the Cauchy-Schwarz inequality
=(√n∑i=1x2i+√n∑i=1y2i)2 =(∥x∥2+∥y∥2)2
Thus we see: ∥x+y∥22≤(∥x∥2+∥y∥2)2, as norms are always ≥0 we see:
∥x+y∥2≤∥x∥2+∥y∥2 - as required.