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Norm vs. a Metric

by Bachelier
Tags: metric, norm
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Bachelier
#1
Feb12-13, 10:55 PM
P: 376
Are the axioms of a Norm different from those of a Metric?

For instance Wikipedia says:

a NORM is a function p: V → R s.t. V is a Vector Space, with the following properties:
For all a ∈ F and all u, v ∈ V, p(av) = |a| p(v), (positive homogeneity or positive scalability).
p(u + v) ≤ p(u) + p(v) (triangle inequality or subadditivity).
If p(v) = 0 then v is the zero vector (separates points).

While a metric is defined as a distance function between elements of a set with the 3 familiar axioms, of positive definition, triangular inequality and symmetry.

The terms are however used loosely in many math books and make you think they are one and the same.

The reason I am asking is I am faced with a question where given a norm definition on the set ##\mathbb{Q}## I am asked to determine if the operation defines a norm? So I was wondering which axioms do I check, those of the Metric or those of the VS norm?

Thank you
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jbunniii
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Feb13-13, 12:17 AM
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A norm and a metric are two different things. The norm is measuring the size of something, and the metric is measuring the distance between two things.

A metric can be defined on any set ##S##. It is simply a function ##d## which assigns a distance (i.e. a non-negative real number) ##d(x,y)## to any two elements ##x,y \in S##. It must satisfy three criteria in order to qualify as a metric. For all ##x,y,z\in S##, (1) ##d(x,y) \geq 0##, and ##d(x,y) = 0## if and only if ##x = y##; (2) ##d(x,y) = d(y,x)##; (3) ##d(x,z) \leq d(x,y) + d(y,z)##.

If ##S## happens to be a vector space, meaning that there is a way to add the elements together and multiply them by scalars, then we can define a norm. This is a function ##\rho## which maps each element to a nonnegative real number. It must satisfy the following rules for all ##x,y \in S## and all scalar ##\alpha##: (1) ##\rho(x) \geq 0##, and ##\rho(x) = 0## if and only if ##x = 0##; (2) ##\rho(\alpha x) = |\alpha| \rho(x)##; (3) ##\rho(x+y) \leq \rho(x) + \rho(y)##.

If we have a vector space with a norm ##\rho##, it is always possible to define a metric in terms of that norm by putting ##d(x,y) = \rho(x-y)##. But not every metric is defined in terms of a norm, or even CAN be defined in terms of a norm. An example is the metric
$$d(x,y) = \begin{cases}
1 & \textrm{ if }x \neq y \\
0 & \textrm{ if }x = y \\
\end{cases}$$
It is easy to check that this satisfies the three conditions for a metric. But even if we define this metric on a vector space, there is no norm ##\rho## satisfying ##d(x,y) = \rho(x-y)##. To see this, suppose there were such a ##\rho##, and let ##\alpha## be any scalar such that ##\alpha \neq 0## and ##|\alpha| \neq 1##. Choose any ##x,y## such that ##x \neq y##. Then ##\alpha x \neq \alpha y##, and we must have
$$1 = d(\alpha x, \alpha y) = \rho(\alpha x - \alpha y) = \rho(\alpha(x-y)) = |\alpha| \rho(x-y) = |\alpha| d(x,y) = |\alpha|$$
which is a contradiction.
Bachelier
#3
Feb13-13, 04:45 AM
P: 376
Quote Quote by jbunniii View Post
A norm and a metric are two different things. The norm is measuring the size of something, and the metric is measuring the distance between two things.

A metric can be defined on any set ##S##. It is simply a function ##d## which assigns a distance (i.e. a non-negative real number) ##d(x,y)## to any two elements ##x,y \in S##. It must satisfy three criteria in order to qualify as a metric. For all ##x,y,z\in S##, (1) ##d(x,y) \geq 0##, and ##d(x,y) = 0## if and only if ##x = y##; (2) ##d(x,y) = d(y,x)##; (3) ##d(x,z) \leq d(x,y) + d(y,z)##.

If ##S## happens to be a vector space, meaning that there is a way to add the elements together and multiply them by scalars, then we can define a norm. This is a function ##\rho## which maps each element to a nonnegative real number. It must satisfy the following rules for all ##x,y \in S## and all scalar ##\alpha##: (1) ##\rho(x) \geq 0##, and ##\rho(x) = 0## if and only if ##x = 0##; (2) ##\rho(\alpha x) = |\alpha| \rho(x)##; (3) ##\rho(x+y) \leq \rho(x) + \rho(y)##.

If we have a vector space with a norm ##\rho##, it is always possible to define a metric in terms of that norm by putting ##d(x,y) = \rho(x-y)##. But not every metric is defined in terms of a norm, or even CAN be defined in terms of a norm. An example is the metric
$$d(x,y) = \begin{cases}
1 & \textrm{ if }x \neq y \\
0 & \textrm{ if }x = y \\
\end{cases}$$
It is easy to check that this satisfies the three conditions for a metric. But even if we define this metric on a vector space, there is no norm ##\rho## satisfying ##d(x,y) = \rho(x-y)##. To see this, suppose there were such a ##\rho##, and let ##\alpha## be any scalar such that ##\alpha \neq 0## and ##|\alpha| \neq 1##. Choose any ##x,y## such that ##x \neq y##. Then ##\alpha x \neq \alpha y##, and we must have
$$1 = d(\alpha x, \alpha y) = \rho(\alpha x - \alpha y) = \rho(\alpha(x-y)) = |\alpha| \rho(x-y) = |\alpha| d(x,y) = |\alpha|$$
which is a contradiction.
Thank you for clarifying this: so which axioms do I have to check for the norm?

HallsofIvy
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Feb13-13, 09:27 AM
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Norm vs. a Metric

Given a norm, |v|, we can define a metric: d(u, v)= |u- v|. However, there exist metric spaces which do NOT correspond to a norm. The simplest is the "discrete metric": for any set, define d(x, y)= 0, if x= y, 1, otherwise.

It is also true that, given any inner product, <u, v>, we can define a norm: [itex]|v|= \sqrt{<v, v>}[/itex], and so a metric. But there exist norms that do not correspond to any inner product.

(Edited thanks to micromass.)
micromass
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Feb13-13, 10:29 AM
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Quote Quote by HallsofIvy View Post
But there exist inner product spaces which cannot be given norms.
I guess you mean this backwards: there exists normed spaces which cannot be given an inner product.
jbunniii
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Feb13-13, 11:45 AM
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Quote Quote by Bachelier View Post
Thank you for clarifying this: so which axioms do I have to check for the norm?
Check that it satisfies the conditions that I mentioned above:
If ##S## happens to be a vector space, meaning that there is a way to add the elements together and multiply them by scalars, then we can define a norm. This is a function ##\rho## which maps each element to a nonnegative real number. It must satisfy the following rules for all ##x,y \in S## and all scalar ##\alpha##: (1) ##\rho(x) \geq 0##, and ##\rho(x) = 0## if and only if ##x = 0##; (2) ##\rho(\alpha x) = |\alpha| \rho(x)##; (3) ##\rho(x+y) \leq \rho(x) + \rho(y)##.
0xDEADBEEF
#7
Feb13-13, 12:36 PM
P: 824
Norms are linear. Metrics only need to fulfil the triangle inequality. If you want to see a cool metric that cannot be derived from a norm look up the French railroad metric.
Fredrik
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Feb13-13, 12:45 PM
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If you're asked to check if a given function is a norm, you must of course (directly or indirectly) check if the function satisfies the conditions listed in the definition of "norm". Anything else would be absurd. If there had been a theorem that says that every metric is a norm, then it would of course have been sufficient to check that the given function is a metric. But there's no such theorem. The conditions that a metric is required to satisfy do not even mention a vector space structure.

Another thing worth noting is that the term "metric" is also used as a short form of "metric tensor field" in differential geometry. That kind of metric is a function that takes each point p in a manifold M to something very similar to an inner product on the tangent space of M at p. (The tangent space at p is a vector space). So I would say that inner products are more closely related to the "metrics" of differential geometry than the metrics of metric spaces.
HallsofIvy
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Feb13-13, 12:52 PM
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Quote Quote by micromass View Post
I guess you mean this backwards: there exists normed spaces which cannot be given an inner product.
Yes, thanks for catching that. I have gone back and edited it.


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