3 ms·
Most of the special properties can be traced back to its special relationship with the inner product. And inner products have somewhat more elementary propertie
by shiandow 1y ago
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 1y 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 https://news.ycombinator.com/item?id=45248881
- JadeNB 1y agoI 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.
- deleted 1y ago[deleted]
- shiandow 1y agoI 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?
- srean 1y agoNot quite sure what you mean but now I agree with your previous point that the coordinate system is irrelevant because these normal derived Lp metrics can be specified in a coordinate free way just by specifying their unit ball set
- shiandow 1y agoWel what does it mean to be orthogonal? I'd argue that if you only have a metric then a line (segment) is orthogonal to some set if it is the shortest path from a point to a set. For norms it should work to only consider straight lines. What I was getting at is that the x and y grid is orthogonal for all Lp norms. The interesting stuff happens if you pick a different line as x-axis. Orthogonal projection to a different axis still works though, the resulting grid just won't be orthogonal in the conventional sense.