Difference between revisions of "Euclidean norm"

Revision as of 03:25, 8 March 2015 (view source) Alec (Talk | contribs) (Created page with "This is an example of a Norm ==Definition== The Euclidean norm is denoted <math>\|\cdot\|_2</math> is a norm on <math>\mathbb{R}^n</math> Here for <math>x\in\mathbb{R}^n...")		Latest revision as of 07:23, 27 April 2015 (view source) Alec (Talk | contribs) m
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Latest revision as of 07:23, 27 April 2015

This is an example of a Norm

Definition

The Euclidean norm is denoted

$$\|\cdot\|_2$$ is a norm on $$\mathbb{R}^n$$

Here for

$$x\in\mathbb{R}^n$$ we have:

$$\|x\|_2=\sqrt{\sum^n_{i=1}x_i^2}$$

Proof that it is a norm

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TODO: proof

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Part 4 - Triangle inequality

Let

$$x,y\in\mathbb{R}^n$$

$$\|x+y\|_2^2=\sum^n_{i=1}(x_i+y_i)^2$$ $$=\sum^n_{i=1}x_i^2+2\sum^n_{i=1}x_iy_i+\sum^n_{i=1}y_i^2$$ $$\le\sum^n_{i=1}x_i^2+2\sqrt{\sum^n_{i=1}x_i^2}\sqrt{\sum^n_{i=1}y_i^2}+\sum^n_{i=1}y_i^2$$ using the Cauchy-Schwarz inequality

$$=\left(\sqrt{\sum^n_{i=1}x_i^2}+\sqrt{\sum^n_{i=1}y_i^2}\right)^2$$ $$=\left(\|x\|_2+\|y\|_2\right)^2$$

Thus we see:

$$\|x+y\|_2^2\le\left(\|x\|_2+\|y\|_2\right)^2$$ , as norms are always $$\ge 0$$ we see:

$$\|x+y\|_2\le\|x\|_2+\|y\|_2$$ - as required.