0
$\begingroup$

I asked a similar question already, but thinking about the problem a bit more I believe I now know better what my actual problem is. In the other question I asked whether the definition of magnitude of vectors could somehow be proved, as it can indeed geometrically be shown to be correct for 1, 2 and 3 dimensions. But proving a definition is kind of nonsensical, therefore I realised what I actually wanted to ask, is how the definition can be proved to be correct. By this distinction I mean, that it needs to be shown, that the definition of magnitude for vectors of any dimensions actually fulfils certain properties of "the concept of length", for example:

A length of the thing times a factor is the length of the thing stretched by that factor:

$c|\vec v| = |c \vec v|$

and

Adding two things and taking their sum's length should be the same as adding the lengths of the things:

$|\vec v + \vec w| = |\vec v| + |\vec w|$

So that would be proving that the function $f : x \mapsto |x| \text{ where } x \in \mathbb{R}^n \text{ and } |x| \in \mathbb{R}$ is linear and a homomorphism.

I guess therefore my real question is, which are the properties that need to be proved to be obeyed by the definition of magnitude to be correct and is there a name for such a function?

To my original question I got the answer that the correctness could be proved by induction, which so far I cannot quite put together with my new realisations of what actually needs to be proved to show that magnitude behaves as one would intuitively expect.

$\endgroup$
2
$\begingroup$

There are various names you could use. A typical one is norm, which axiomatizes properties that a reasonable notion of the length of a vector should have. The axioms are

  1. $|v| \ge 0$ for all $v$, and if $|v| = 0$ then $v = 0$.
  2. $|cv| = |c| |v|$, where $c$ is a scalar and $v$ is a vector.
  3. $|v + w| \le |v| + |w|$, the triangle inequality.

None of these are difficult to prove for the $\ell^2$ or Euclidean norm

$$|(x_1, \dots x_n)| = \sqrt{x_1^2 + \dots + x_n^2}$$

although the triangle inequality takes a bit of work.

But it's unclear to me whether this captures what you mean by "correct" (which I don't yet understand). Among other things, this definition is satisfied by many notions of norm other than the usual Euclidean one, for example the $\ell^1$ or "taxicab" or "Manhattan" norm

$$|(x_1, \dots x_n)| = |x_1| + \dots + |x_n|.$$

Perhaps you are looking for a set of properties which uniquely characterize the Euclidean norm.

$\endgroup$
1
  • $\begingroup$ I think this mostly clears it up. The only confusion remaining is the relation between proving that the properties hold, and deriving the formula for n dimensions from the formula for 2 dimensions via induction. Does the proof by induction already imply that the necessary properties hold, or why would one do it (the proof by induction)? $\endgroup$ – user3578468 Sep 27 '17 at 9:44
1
$\begingroup$

Unfortunately, your requirements that $c|{v}| = |cv|$ and $|v+w|=|v|+|w|$ are not true for our usual definition of the magnitude in $\mathbb{R}^n$. Instead, with the usual definition it is easy to prove that $|c| \cdot |v| = |cv|$ and $|v+w| \leq |v| + |w|$ hold, both desirable properties of a magnitude definition.

$\endgroup$
2
  • $\begingroup$ Ah right yes, triangle inequality. But which are the other properties that need to hold? $\endgroup$ – user3578468 Sep 26 '17 at 16:04
  • $\begingroup$ Maybe it has something to do with the dot product. In an inner product space (like R^n with the dot product), you define the magnitude of a vector v to be the nonnegative square root of the inner product of v with itself. If you do that then the magnitude formula is forced. $\endgroup$ – Leonard Blackburn Sep 26 '17 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.