I am doing machine learning and in one stage, I have to measure tensor vector differences to find the minimum distance. Lets say I have a set of linear tensors with dimension 100. I want to find a best pair, which in this case is a pair with minimum distances from each other. So, I have to find two tensors a and b that are closest ones to each other i.e. d(a,b) < d(c,d) for any tensors c and d in our set. I wonder if using these two distance measurements give same results or not? And if they are not, which one is better? In terms of speed, I guess using L1 is better. So, is L1 as reliable as L2, making it better one?
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$\begingroup$ Just to add some more precision in the question: a and b are real vectors of dimension n>2? Are you interested in a statistical statement where a, b, c, d follow some distribution or are you interested in a statement for a certain fixed realization of these? $\endgroup$– Ggjj11Commented Oct 11, 2023 at 20:30
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$\begingroup$ I added more description to my question. $\endgroup$– AliMCommented Oct 11, 2023 at 21:21
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$\begingroup$ Could you please explain what you mean by "L1" and "L2"? Are these perhaps referring to $L_1$ and $L_2$ distances relative to some basis for the tensors? I also wonder about the phrase "linear tensors," because part of the usual definition of a tensor is that it is (or represents) a multilinear map. What would a "nonlinear tensor" be? $\endgroup$– whuber ♦Commented Oct 12, 2023 at 11:48
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$\begingroup$ whuber, I got my answer. For L1 and L2, I meant Manhatan and Euclidean distances and also by linear tensor, I meant like vectors and not matrices. Thanks anyway. $\endgroup$– AliMCommented Oct 12, 2023 at 17:25
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$\begingroup$ I am glad you have an answer. But if future visitors cannot determine what your question is asking, then the entire thread can be confusing or wrong, depending on how it is interpreted. BTW, matrices are vectors. $\endgroup$– whuber ♦Commented Oct 13, 2023 at 14:28
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1 Answer
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No.
Consider $a = [0,1], b = [0,2], c=[0,1], d=[0.51,1.51]$. $L^1(a,b)=1 < L^1(c,d) = 1.02, L^2(a,b)= 1 >L^2(c,d) = 0.721...$, reversing the direction of the inequality.
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$\begingroup$ Thanks jbowman. Now that they are not equivalent, which one is more reliable for computing distance? Is it L2? $\endgroup$– AliMCommented Oct 11, 2023 at 21:45
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1$\begingroup$ L2 might be more sensitive to deviations in one of the components... $\endgroup$– Ggjj11Commented Oct 11, 2023 at 21:47
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1$\begingroup$ Is there a connection to the unit circles of pnorms? commons.m.wikimedia.org/wiki/File:Vector-p-Norms_qtl1.svg could you work this out? $\endgroup$– Ggjj11Commented Oct 11, 2023 at 22:04
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$\begingroup$ thanks Ggjj11 for your comments. I got the answer. Your comments were very helpful $\endgroup$– AliMCommented Oct 12, 2023 at 17:28
