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I've been re-reading my linear algebra book and a definition is given of the norm of a vector in $\mathbb{R}^n$ to be:

||v|| $= \sqrt{v_1^2 + v_2^2 + \cdots + v_n^2}$.

For $\mathbb{R}^2$ and $\mathbb{R}^3$ this can be trivially verified by applying the Pythagorean theorem. I struggle, however, to see how this is a definition we can assert to be true in $n$ space.

I think I can see a way that this could be proven inductively, but I'm not sure if induction even really makes sense to use here. The way I see it, when we extend the case from $\mathbb{R}^2$ to $\mathbb{R}^3$, we essentially create one leg of a right triangle in the $xy$ plane by projecting our $z$ coordinate orthogonally into the plane and finding that distance, while the other leg is precisely the length of that orthogonal projection of the $z$ component onto its shadow in $\mathbb{R}^2$. If this line of thinking is correct, couldn't we could find the length of a vector in $\mathbb{R}^4$ by again projecting the fourth component, let's call it $w$ into $\mathbb{R}^3$, where that distance becomes one leg of our right triangle.

Then, since this newly added fourth dimension introduces a fourth orthogonal axis, we know our initial projection was orthogonal and then the length of the other leg of this (hyper?) triangle is precisely the $w$ component that was projected, thus from this we are applying some type of four dimensional case of the pythagorean theorem to find our vector's norm. If we accept that approach as valid, for any vector of length $n$ we could continually construct the norm of that vector by building up to the length in dimension $n-1$ and the $n$th component consequentially will be our other leg of the (hyper?) triangle. That would be because the distance of the orthogonal projection of the $n$th component into $n-1$ space would again be precisely the value of said $n$th component.

I am trying to find a way to relate what the definition of the Euclidean norm is saying to something constructive and concrete. Has what I said made any sense at all, or does this come from a failure to understand the way we extend the notion of "distance" in higher dimensions?

If I had to summarize my question, it would be: "Is the Euclidean norm simply an agreed-upon convention to define distance in $n$ space that assumes the way we calculate magnitude is dimension invariant, or is there a more concrete way to prove that this definition really is tied to something concrete and can be rigorously verified?"

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    $\begingroup$ We don't "verify" for $\Bbb R^2$ and $\Bbb R^3$ using Pythagorean theorem, we define the norm like that, for $\Bbb R^2$, for $\Bbb R^3$ and in general, for $\Bbb R^n$. There's nothing to prove about it, since is a definition. $\endgroup$
    – jjagmath
    Commented Jun 22 at 1:43
  • $\begingroup$ @jjagmath perfect, that's all I needed to hear to be honest. I was trying to reconcile it in my head as if it was something that could / had to be proven. $\endgroup$
    – MattKuehr
    Commented Jun 22 at 1:53
  • $\begingroup$ You can use the pythagorean theorem to prove the diagonal of a rectangle has length $\sqrt{x^2+y^2+z^2}$ geometrically. The same algebraic manipulation you use for that case can be extended to $\mathbb{R}^n$. You can skip this by taking it as a definition but I think it's instructive to motivate the definition in this way. $\endgroup$ Commented Jun 22 at 1:56

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This is subtle. In mathematics we have the freedom to define distances however we want depending on the application, and for example sometimes rather than the Euclidean norm we use what are called the $\ell^p$ norms

$$\| x - y \|_p = \sqrt[p]{\sum_{i=1}^n |x_i - y_i|^p}$$

and even stranger options are possible.

So the question arises: what is special about the Euclidean norm in higher dimensions, among all the other choices we have for measuring distances in $\mathbb{R}^n$? After all, we can justify the Euclidean norms in dimensions $2$ and $3$ by arguing that they (approximately) describe the geometry of the $3$-dimensional space we inhabit and the $2$-dimensional planes inside it, but no such argument is available in dimensions $n \ge 4$.

One way to answer this question is the following, which is rarely stated but which I feel like deserves to be better known.

Claim 1: Up to a change of coordinates, the Euclidean norm is the unique vector space norm on $\mathbb{R}^n, n \ge 3$ with the property that it restricts to the Euclidean norm on every plane through the origin in $\mathbb{R}^n$, after a suitable choice of coordinates.

This is a consequence of the polarization identity, which allows us to recover the dot product from the Euclidean norm. When the polarization identity is applied to any norm other than a Euclidean norm (up to change of coordinates) it fails to produce a function that behaves like the dot product should, and moreover this condition can be checked on planes.

Another very special feature of the Euclidean norm is the following.

Claim 2: Again up to a change of coordinates, the Euclidean norm is the unique vector space norm on $\mathbb{R}^n, n \ge 2$ with the property that its isometry group acts transitively on the unit sphere $\{ x \in \mathbb{R}^n : \| x \| = 1 \}$.

In less technical language, every norm defines a collection of "generalized rotations and reflections" which share the same property that rotations and reflections do that they don't change the distances between vectors (this is what "isometry" means). Every norm also defines a "generalized unit sphere" which consists of all vectors at a distance of exactly $1$ from the origin. The above says that the Euclidean norm, which produces the usual unit sphere, has the "most symmetric" unit sphere in the sense that every unit vector can be rotated or reflected to become every other unit vector. For a proof you can see here.

Loosely speaking this says that the Euclidean norm is the "least biased" norm in the sense that it does not prefer any direction to any other (it is "isotropic"). Every other choice of norm makes some directions "special" compared to others. You can sort of get a visual sense for this by looking at, for example, this lovely visualization of the unit balls of the $\ell^p$ norms in dimension $3$.


Edit: Here is a completely different characterization of the Euclidean norm which I find very suggestive but also mysterious. Consider a random walk on the integer lattice $\mathbb{Z}^n$, where we start at the origin and at every time step we walk one step in each coordinate direction, either $+1$ or $-1$ (so there are $2^n$ possible steps and each has equal probability $\frac{1}{2^n}$). For every time $t$ and every point in $\mathbb{Z}^n$ we can consider the probability that the random walk hits that point after that time. A priori the only symmetry of this situation is the freedom to permute and negate the coordinate axes, and this is a finite collection of symmetries.

Now consider rescaling this random walk; that is, zoom out so that the integer lattice $\mathbb{Z}^n$ looks like a smaller lattice, say $\frac{1}{r} \mathbb{Z}^n$.

Claim 3: As $r, t \to \infty$, the rescaled random walk has a probability distribution in which the probability of hitting a point depends only on the Euclidean distance of that point from the origin.

This follows from the central limit theorem, and in fact this rescaled random walk converges in a suitable sense to Brownian motion (this is Donsker's theorem). Among other things this means the rescaled random walk has much more symmetry than the original random walk, namely the entire orthogonal group of rotations and reflections. It really makes you wonder what this result has to do with the fact that the physical space we inhabit is approximately Euclidean...

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    $\begingroup$ Personally I find this stuff very mysterious. Euclidean geometry is quite special in a way I'm not sure I can really claim to understand. $\endgroup$ Commented Jun 22 at 3:28
  • $\begingroup$ Wow. I can't see how people with non-$\ell^2$ "eyes" would be able to "see" those preferred directions (which I initially assumed were the corners, but my first pass though the proof you linked makes me think might be the centers of the faces). I'm going to have to spend some time going over that proof. Thanks for posting this answer! $\endgroup$
    – JonathanZ
    Commented Jun 22 at 3:48
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    $\begingroup$ @JonathanZ: right, the visualization is a little misleading because of course it's not as if people who are "intrinsically $\ell^p$" can see the $\ell^2$ norms of anything. I am just referring to the fact that the isometry group is much smaller (for the $\ell^p$ norms where $p \neq 2$ it is finite) and so there are many more orbits of the isometry group acting on the unit sphere. This suggests that, for example, triangles oriented differently could behave pretty differently, things like that. The classification of "equilateral triangles" ought to be pretty sensitive to direction. $\endgroup$ Commented Jun 22 at 3:50
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    $\begingroup$ Something I find interesting is that in the $p$-adic world, the analogues of these special properties are satisfied by the sup-norm $\ell^\infty$ instead of $\ell^2$, so the most symmetric $p$-adic spheres are actually the cubes! (I described some of the analogies implicitly in my question here). $\endgroup$
    – pregunton
    Commented Jun 23 at 10:31
  • $\begingroup$ @pregunton: wow, that's an interesting observation in your question. I hope someone gives a good answer! I've offered a bounty. $\endgroup$ Commented Jun 23 at 18:19
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The definition is partly "an greed-upon convention", but at the same tame there is a reason to define it the way it is.

There is mathematical notion called "metric" or "distance". Given a set $X$, a metric $d$ is any function $d : X \times X \to \mathbb{R}$ that satisfies:

(I) $d(x,x) = 0$ for any $x \in X$ (the distance from a point to itself is always $0$).

(II) $d(x,y) = d(x,y)$ for any $x,y \in X$ (the distance from a point $x$ to another $y$ is the same if you measure it from point $y$ to the point $x$).

(III) $d(x,y) \leq d(x,z) + d(z,y)$ for any $x,y,z \in X$ (this one is a bit tricky, it's called the triangle inequality, it basically says that the distance from point $x$ to point $y$ is always less or equal than the distance from $x$ to any intermediate point $z$, and the distance from that $z$ to the end point $y$).

$(X,d)$ is called a metric space, and is an extremely important definition in mathematics, it basically captures the notion of having a criteria of "distance" for any set.

Notice that, in the particular case for $\mathbb{R}^2$, if we define:

$$d_{\mathbb{R}^2}(x,y) := \sqrt{(x_1-y_1)^2 + (x_2 - y_2)^2}$$

where $x = (x_1,x_2)$ and $y = (y_1,y_2)$, then $d_{\mathbb{R}^2}(x,y)$ is a metric in $\mathbb{R}^2$! (it fulfills I,II and III), in layman terms "$d_{\mathbb{R}^2}$ is a notion of distance I can use for $\mathbb{R}^2$".

Now, for $\mathbb{R}^3$, if we define:

$$d_{\mathbb{R}^3}(x,y) := \sqrt{(x_1-y_1)^2 + (x_2 - y_2)^2 + (x_3 -y_3)^2}$$

where $x = (x_1,x_2,x_3)$ and $y = (y_1,y_2,y_3)$, then $d_{\mathbb{R}^3}$ is also a metric in $\mathbb{R}^3$.

Now you can see where this is going, right?

For $\mathbb{R}^n$, $n > 3$, we define:

$$d_{\mathbb{R}^n}(x,y) := \sqrt{\sum_{i=1}^n(x_i-y_i)^2}$$

where $x = (x_1,x_2,x_3,...,x_n)$ and $y = (y_1,y_2,y_3,...,y_n),$ then YOU CAN PROVE that $d_{\mathbb{R}^n}$ is also a metric in $\mathbb{R}^n$!!!

Final observation: what does this have to do with norms?, well, notice that $d_{\mathbb{R}^n}(x,y) = \|x - y\|_{\mathbb{R}^n}$. You can prove that any normed space is a metric space, but not every metric can be induced by a norm.

So, basically, yeah is "an greed-upon convention", but at the same tame you can see that this definition agrees with the abstract notion of distance (metric), and even more, it is also a norm (a way to asign a "magnitude" to every point in the set). I'm not writing here the abstract general definition of a norm, but you can check it out here: https://en.wikipedia.org/wiki/Norm_(mathematics).

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