TITLE: The difference between $L_1$ and $L_2$ norm?
QUESTION [6 upvotes]: I have been trying to understand what is the difference between $L_1$ and $L_2$ norm and cant figure it out.
In this webpage I got a clear understanding of why we would use $L_1$ norm (scroll down till you get to the          google maps picture).
I went on matlab and calculated the norm for the matrix $A=[3, 7]$ and got that the $L_1$ norm is $10$, which makes sense as in the example above. It is the distance between $(0,0)$ and $(3,7)$. When I do the $L_2$ norm I get $7.61\dots$ and then $L_3$ is $7.1$ and so on until it converges to $7$. What do these calculations mean? Why are the numbers getting smaller and converging to $7$. In my intuition I can grasp why would the distance between $(0,0)$ and $(3,7)$ be $10$, but cant understand the need of $L_2$ and that being $7.61$. I looked everywhere for an intituitive explanation but all I get is how to calculate the norm, which I already know how to do.
Any reason why would it be best to use $L_2$ to calculate the distance-magnitude of a vector and why is the number smaller in $L_2$ than the intituitive case of $L_1$? Thanks.

REPLY [3 votes]: The $L^2$ norm is a straightforward generalization of the standard Euclidean norm on $\mathbb{R}^n$.
You can ask why $L^2$ norm is so good, or why it is so popular?
The answer is simple: the $L^2$ norm is easiest to play with, it is easiest to perform manipulations with it, and what is perhaps the most important it is easy to calculate.
All that good properties of that norm have one simple source: the $L^2$ norm is a norm induced by an inner product on a Hilbert space -$L^2$ is a Hilbert space. And when you are dealing with a Hilbert space you can do almost anything, you have orthonormal bases, orthogonal projections and other Hilbert's space gadgetery. This norm comes with a rich structure that enables you to do many things with ease. That is the true rationale standing behind the popularity of the $L^2$ norm in applications.