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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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    The answer below contains the type of link I was looking for: http://en.wikipedia.org/wiki/Norm_%28mathematics%29#Definition2012-10-13

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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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    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.2012-10-12
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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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    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.2012-10-12
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The 2-norm $ \left|v\right| = \sqrt{\left} $ 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.