# Intuition of Euclidean norm

I know the following :

A norm is Euclidean iff it respects the parallelogram law.

The problem is that the parallelogram law is not very intuitive geometrically. So I am wondering if there is a geometric intuition about Euclidean norms. This intuition should help being able to distinguish Euclidean norms from non-Euclidean ones.

Thank you!

• “The problem is that the parallelogram law is not very intutitive geometrically.” Have you seen the picture behind it? – Gunnar Þór Magnússon Feb 7 at 13:16
• @TodorMarkov An euclidiean norm is norm induced by an inner product. – dghkgfzyukz Feb 7 at 13:30
• @GunnarÞórMagnússon yes I've seen the picture behind but the problem is that it relies on the pythagorian theorem which is not intuitive... That's why I am seeking for a simple intuition but maybe there isn't... – dghkgfzyukz Feb 7 at 13:31
• If you visualize a vector $x \in \mathbb R^n$ as an arrow, the Euclidean norm is just the length of that arrow. That seems like a very intuitive way to measure the size of a vector. – littleO Feb 8 at 1:32

Let's first look at a vector space without norm. Such a vector space has one special vector, the zero vector, but other than that, all vectors are essentially equivalent: There's no way to distinguish one from the other just by using the operations of a vector space. There's absolutely no property of any non-zero vector that any other non-zero vector does not have as well. Now, of course if you are given two vectors, you can tell whether they are the same, or whether they are linearly dependent, and if so, with which factor; however if they are independent, that's again all you can say about them.

Now why am I telling you that? Well, because that radically changes as soon as we have a norm. The norm allows you to split any vector $$v$$ into its length $$\|v\|$$ and its direction $$\frac{v}{\|v\|}$$. So you can classify vectors according to their length. Well, still kind of obvious.

But we can now ask: Are at least the different directions still essentially the same? Or are there any preferred directions that you can identify just with the operations of a normed vector space?

Well, for a general norm, the directions are indeed not equal. For example, take the maximum-norm on $$\mathbb R^2$$, given by $$\left\|\pmatrix{x\\y}\right\| = \max\{|x|,|y|\}$$ and consider the four vectors $$a=\pmatrix{1\\0}, b=\pmatrix{0\\1}, c=\pmatrix{1\\1} \text{ and } d=\pmatrix{1\\-1}.$$ It is easily checked that each of them is an unit vector under that norm. However they are easily distinguished: For the vector $$a$$, we habe $$a+b=c$$ and $$a-b=d$$, that is, there exists an unit vector, $$b$$, for which both adding and subtracting each gives yet another unit vector. The same is, of course, true for $$b$$. On the other hand, neither $$c$$ nor $$d$$ allow such an unit vector. So this norm clearly allows to distinguish $$a$$ from $$c$$, by the existence of such a vector.

Now this is special to that specific norm; most other norms won't allow this specific way of distinguishing.

However let's look specifically at the quantity $$f(v) := \min_{\|w\|=1}g(v,w),\quad g(v,w)=(\|v+w\|^2+\|v-w\|^2).$$ We can easily check that $$g(v,\frac{v}{\|v\|}) = (\|v\|+1)^2 + (\|v\|-1)^2 = 2\|v\|^2 + 2$$ and thus $$f(v)\le 2\|v\|^2+2$$, and in particular for unit vectors, $$f(v)\le 4$$.

Let's check this quantity on our maximum-norm example from above. A straightforward calculation gives $$f(a)=2, f(c)=4$$. So we see that this function indeed allows us to distinguish the different directions.

On the other hand, one readily checks that if the parallelogram law holds, then for any vector $$v$$, we have $$f(v)=2\|v\|+2$$, that is, $$f$$ does not allow to distinguish between different vectors of the same norm.

Indeed, I think the converse should also hold: A norm that doesn't distinguish directions (that is, $$f(v)$$ only depends on $$\|v\|$$) necessarily fulfils the parallelogram law. Which would mean that the Euclidean norm is the unique norm that treats all directions equally. However at the moment I don't see how to show it.

Edit:

It seams I haven't been as clear as I thought I was, therefore I'll clarify.

Let me first formally define what I mean when I say that an algebraic structure doesn't distinguish certain objects. Explaining that was the point of my first paragraph above, but judging from the comment, I failed. Therefore here the formal definition:

An algebraic structure does distinguish between objects $$a$$ and $$b$$ if there is a statement $$\phi(x)$$, using only the tools given by the algebraic structure, such that $$\phi(a)$$ is true and $$\phi(b)$$ is false.

Note that $$\phi(x)$$ has $$x$$ as its only free variable, and in particular does not contain any constant object from the algebraic structure unless that object can itself be specified using only the tools of the algebraic structure itself (and of course those provided by logic itself, in particular equality).

For example, a vector space without norm does distinguish the zero vector from any non-zero vector using e.g. the statement $$\phi(x) \equiv x + x = x$$. That statement clearly is true for for the zero vector, and is false for any other vector. And that statement only uses the tools provided by the algebraic structure (here specifically, addition).

On the other hand, a vector space without norm does not distinguish any two non-zero vectors. That is, given two non-zero vectors $$a$$ and $$b$$, there is no formula $$\phi(x)$$ such that $$\phi(a)$$ is true and $$\phi(b)$$ is false.

Now you might say that e.g. in $$\mathbb R^2$$, the formula $$\phi(x) \equiv x=\pmatrix{1\\0}$$ is such a formula. But note that the vector $$\pmatrix{1\\0}$$ is not provided by the vector space structure. That is, there is nothing in the vector space axioms that tells us that the vector space contains a vector $$\pmatrix{0\\1}$$.

You might also say that $$\phi(x) \equiv x=a$$ is a formula that distinguishes $$a$$ from $$b$$, as clearly it is true for $$a$$ and false for $$b$$. But again, $$a$$ is not supplied by the vector space axioms, and to allow it to be supplied at the same time as $$b$$, we'd need two free variables, that is, $$\phi(x,y)\equiv x=y$$, so that we then can set $$y$$ to $$a$$. But the definition above allows only for one free variable.

OK, so after having our definition, let's consider a normed vector space $$V$$. As I wrote above, now the non-zero vectors are no longer all equal; for example, if we choose $$\mathbb R^2$$ with the maximum-norm, then we can distinguish between the vector $$a=\pmatrix{1\\0}$$ and the vector $$b=\pmatrix{2\\0}$$ e.g. through the statement $$\phi(x) \equiv \|x\| = 1$$, which obviously is true for $$x=a$$, but is false for $$x=b$$.

Side remark: You might doubt whether we may use the constant 1 here. We may, since the norm by definition gives a real number, and the constant $$1$$ is provided to us from the structure of the real numbers (as the identity of multiplication). The same is true for all rational numbers (e.g. $$4$$ is the unique real number $$x$$ such that $$x=1+1+1+1$$) and for many irrational numbers as well. Indeed, you can be assured that any real number that can be explicitly written can also be distinguished from any other real number.

Now, as I wrote above, the norm allows you to split any non-zero vector $$v$$ into a length (its norm $$\|v\|$$) and a direction (the unit vector $$n_v$$ for which $$\|v\|n_v = v$$, that is, $$n=v/\|v\|$$). And as we have seen, vectors of different norm can always be distinguished.

So what if the norm of both vectors cannot be distinguished from all others in $$\mathbb R$$? Well, between two arbitrary real numbers, there's always a rational number $$q$$, and that one we can distinguish from any other. And then we can use $$\phi(x) \equiv \|x\| to distinguish both vectors.

So the remaining question is whether we can distinguish vectors of the same norm. Note that this is equivalent to the question whether we can distinguish directions, since vectors of the same norm only differ in their direction.

Now imagine that we can somehow construct an arbitrary function $$h:V\to \mathbb R$$, such that for two vectors $$a$$, $$b$$ of the same norm, we have $$h(a)\ne h(b)$$. Then we clearly can distinguish $$a$$ from $$b$$. Which in turn means that if a norm cannot distinguish vectors of the same norm, then all functions of the form which you can construct using it will only depend on the norm of the vector.

So what is special about the function $$f$$ I introduced above? Well, on one hand it is a function of the form above, so if for any two vectors $$\|a\|=\|b\|$$, we find $$f(a)\ne f(b)$$, we have proved that the corresponding norm does distinguish directions. And that is just what I demonstrated with the maximum norm.

On the other hand, I constructed it explicitly so that for any norm that respects the parallelogram law, it only depends on the norm. So it is easy to see that for any norm that respects the parallelogram law, this specific function cannot be used to distinguish between vectors of the same norm.

My conjecture is that it is exactly for those norms that respect the parallelogram law that $$f$$ only depends on the norm. Since it is exactly the Euclidean norms that respect the parallelogram law, this would mean that every non-Euclidean norm allows to distinguish different directions.

Now this alone does not suffice to show that the Euclidean norm does not allow to distinguish directions. After all, just because we cannot distinguish directions using $$f$$ doesn't necessarily mean we cannot find another function that distinguishes them. However showing that an Euclidean norm doesn't distinguish directions can be done independently (basically, construct a norm-preserving linear transformation that transforms one of the two vectors into the other one).

One note to the claim in the comment that the inner product allows to distinguish directions: No, it doesn't. It allows to distinguish relative directions (that is, angles between two vectors), but is doesn't distinguish directions as such (that is, given a single vector, there's no way to use the inner product to determine which direction it points to). Indeed, given an Euclidean norm, you can construct the inner product. And of course given the inner product, you can construct the Euclidean norm. That is, the Euclidean norm and the inner product give you exactly the same information.

• (+1), This is really interesting ! It's actually quite strange that Euclidean norms are the one which don't distinguish direction since by definition an Euclidean norm inherited from an inner product. And inner product clearly indicates direction. Moreover I understand the $f$ function, but I don't see the link between $f$ and direction, why $f$ has anything to do with direction ? Thank you ! – dghkgfzyukz Feb 12 at 8:13
• @dghkgfzyukz: See my edit; I hope now it's more clear. – celtschk Feb 13 at 20:43
• Thank you very much, everything is clear now ! That's an interesting conjecture, I'll think about it. – dghkgfzyukz Feb 14 at 8:00

The only answer I know is that the parallelogram law is the geometric intuition for Euclidean norms. If we look at vectors in this picture, we'd really like the parallelogram identity $$2 \|u\|^2 + 2\|v\|^2 = \|u + v\|^2 + \|u - v\|^2$$ to be true for them. However, that identity, and our geometric intuition, only holds for some norms.

Maybe it helps to see some examples of norms. Let's look at a few ones on $$\mathbb R^2$$:

• $$\| (x, y) \|_1 := |x| + |x|$$
• $$\| (x, y) \|_2 := \sqrt{x^2 + y^2}$$
• $$\| (x, y) \|_{\infty} := \max\{|x|, |y|\}$$

The middle one is the usual Euclidean norm we all love. The first one is the "Manhattan" or "taxi cab" distance norm, and the third one is the supremum norm.

Let's now pick some vectors, like $$u = (1,2)$$ and $$v = (3, 4)$$. The parallelogram law says that $$2 \|u\|^2 + 2\|v\|^2 = \|u + v\|^2 + \|u - v\|^2$$. Plugging our vectors and norms into this equation gives:

• $$400 \not= 196$$ for the Manhattan norm
• $$60 = 60$$ for the Euclidean norm
• $$144 \not= 64$$ for the supremum norm

So there is something special about the Euclidean norm. This special thing is that it is the only one that's induced by an inner product; one can easily verify that such a norm satisfies the parallelogram law by using the identity $$\|u + v\|^2 = \|u\|^2 + 2 \langle u, v \rangle + \|v\|^2$$, and when we have a norm that satisfies that law the polarization identities define an honest inner product that induces the norm.

I'm not really sure what else you're looking for.