# Norm of an inner product space

I'm wanting to know why the norm of an inner product space is defined by

$$\|v\| = \langle v | v \rangle^{1/2}$$

I would assume it's not arbitrary, but I don't see anything that would lead it to be defined in this way. What would the consequences be if this definition was changed?

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Your question is unclear. On an inner product space, there is a priori no norm. You can define a norm by that formula. – wildildildlife Oct 12 '12 at 22:53
I guess what I'm getting it as this: why is it defined that way? – user1520427 Oct 12 '12 at 22:54
You want ordinary Euclidean distance as a special case, which you don't get if you use a different definition. – Qiaochu Yuan Oct 12 '12 at 23:05
In one word: Pythagorean theorem. – Berci Oct 12 '12 at 23:54
Because is satisfies the requirements of a norm: $\|u\| \geq 0$ for all $u$ and $=0$ iff $u = 0$, $\|au\| = |a| \|u\|$ for all "scalars" $a$ and "vectors" $u$, and (triangle inequality) $\|u + v\| \leq \|u\| + \|v\|$ for all $u, v$ in the inner product space. I think those are all the properties of a norm - it is surprisingly difficult to find a self-contained definition on the Web. – Stefan Smith Oct 13 '12 at 1:25

The 2-norm $$\left|v\right| = \sqrt{\left<v\big| v\right>}$$ is the most common because it yields the length of a vector $v \in R^n$, with the inner product being the dot product. Any norm satisfying the definition is valid, however, such as the more general p-norm.

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One reason (but probably not main) is that if it wasn't this way it would not always agree with the euclidean conception of distance. Imagine you are asked for the points within a certain distance from one given point and such that the three of them are on a straight line. Intuitively, there are two solutions (each one at "both sides" of the given point). If the norm was the inner product raised to one, then the answer would be a singular point and while this might look OK in an abstract sense, it would not agree with "reality", if you want to call it that way. I mean, you would be missing half the solution to the problem.

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Ok cool, so abstractly there's no real reason to define it that way, but in doing so it agrees with Euclidean distance as Qiaochu Yuan said? – user1520427 Oct 12 '12 at 23:16
Yeah I think so, but still, from the point of view of Functional Analysis there are still powerful reasons to define it that way. As somebody has already commented, if $|| \cdot ||$ wants to be a norm it must retain the properties of a norm. – busman Oct 12 '12 at 23:18

What you wrote is usually the definition of $\|v\|$. If you want it to be derived, what is your definition of $\|v\|$?

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That's a good point and I guess it's a weird question to ask. But what I'm trying to get at is why do we use $\langle v | v \rangle ^ {1/2}$ and not, I don't know, $\langle v | v \rangle ^ 1$ – user1520427 Oct 12 '12 at 22:53
@user1520427: if you don't take the square root, it does not define a norm. – wildildildlife Oct 12 '12 at 23:13
Yeah but that's pretty circular reasoning.. – user1520427 Oct 12 '12 at 23:14
No, it's not circular reasoning. If you try to define a norm by $\|v\|=\langle v,v\rangle$, you get that $\|2v\|=4\|v\|$. As the idea behind a norm is that it is a distance, you expect it to respect scaling, i.e. $\|2v\|$ should be $2\|v\|$. And it doesn't satisfy the triangle inequality either. So, as @wildildildlife said, it is not a norm. – Martin Argerami Oct 12 '12 at 23:28