Comment by shiandow

Comment by shiandow 17 hours ago

4 replies

Most of the special properties can be traced back to its special relationship with the inner product. And inner products have somewhat more elementary properties, so in that sense it explains the special position of the euclidean norm.

This has nothing to do with the coordinates by the way. If you want a different norm you'll first have to figure out an alternative to the bilinearity that gives the inner product its special properties.

Though bilinearity is pretty special itself, given the link between the tensor space and the linear algebra equivalent of currying.

srean 16 hours ago

> This has nothing to do with the coordinates by the way.

I think it does. Both decompose along orthogonal directions. See my comment here https://news.ycombinator.com/item?id=45248881

  • shiandow 5 hours ago

    I mean they decompose in orthogonal components for all Lp norms I think? Is there a norm for which (x,0) is not the closest point to (x,y) on the x-axis?

  • JadeNB 14 hours ago

    I think that's arguably an a posteriori explanation: you can find orthogonal coordinates with respect to which the L^2 norm has a nice form, but you can also single out the L^2 norm in various ways (for example, by its large symmetry group, or the fact that it obeys the parallelogram law—or even just the fact that "orthogonal" makes sense!) without ever directly referencing coordinates.

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