I was just thinking back to this, and it occurred to me that one possible reason to be unsatisfied with the arguments I presented here is that I started off with this notion of a crossing-over point as p continuously increases. But then when you asked “ok, but why is the crossing-over point 2?”, I was like “uh, consider that it might be an integer, and then do a bunch of very discrete-looking arguments that end up showing there’s something special about 2″, which doesn’t connect very well with the “crossover point when p continuously varies” picture. If indeed this seemed unsatisfying to you, then perhaps you’ll like this more:
If we have a norm on a vector space, then it induces a norm on its dual space, given by |φ|:=max|v|=1|φ(v)|. If a linear map preserves a norm, then its adjoint preserves the induced norm on the dual space.
Claim: The Lp norm on column vectors induces, as its dual, the Lq norm on row vectors, where p and q satisfy 1p+1q=1.
Thus if a matrix preserves Lp norm, then its adjoint preserves Lq norm. When p=2, we get that its adjoint preserves the same norm. This sort of gives you a natural way of seeing 2 as halfway between 1 and infinity, and giving, for every p, a corresponding q that is equally far away from the middle in the other direction, in the appropriate sense.
Proof of claim: Given p and q such that 1p+1q=1, and a row vector φ=(φ1,...,φn) with Lq norm 1, let xi=|φi|q, so that x1+...+xn=1. Then let vi:=±x1/pi (with the same sign as φi). The column vector v=(v1,...,vn)T has Lp norm 1. φv=φ1v1+...+φnvn=x1p+1q1+...+x1p+1qn=1. This shows that the dual-Lp norm of φ is at least 1. Standard constrained optimization techniques will verify that this v maximizes φv subject to the constraint that v has Lp norm 1, and thus that the dual-Lp norm of φ is exactly 1.
Corollary: If a matrix preserves Lp norm for any p≠2, then it is a permutation matrix (up to flipping the signs of some of its entries).
Proof: Let q be such that 1p+1q=1. The columns of the matrix each have Lp norm 1, so the whole matrix has Lp norm n1/p (since the entries from each of the n columns contribute 1 to the sum). By the same reasoning about its adjoint, the matrix has Lq norm n1/q. Assume wlog p<q. Lq norm is ≤ Lp norm for q>p, with equality only on scalar multiples of basis vectors. So if any column of the matrix isn’t a basis vector (up to sign), then its Lq norm is less than 1; meanwhile, all the columns have Lq norm at most 1, so this would mean that the Lq norm of the whole matrix is strictly less than n1/q, contradicting the argument about its adjoint.
I was just thinking back to this, and it occurred to me that one possible reason to be unsatisfied with the arguments I presented here is that I started off with this notion of a crossing-over point as p continuously increases. But then when you asked “ok, but why is the crossing-over point 2?”, I was like “uh, consider that it might be an integer, and then do a bunch of very discrete-looking arguments that end up showing there’s something special about 2″, which doesn’t connect very well with the “crossover point when p continuously varies” picture. If indeed this seemed unsatisfying to you, then perhaps you’ll like this more:
If we have a norm on a vector space, then it induces a norm on its dual space, given by |φ|:=max|v|=1|φ(v)|. If a linear map preserves a norm, then its adjoint preserves the induced norm on the dual space.
Claim: The Lp norm on column vectors induces, as its dual, the Lq norm on row vectors, where p and q satisfy 1p+1q=1.
Thus if a matrix preserves Lp norm, then its adjoint preserves Lq norm. When p=2, we get that its adjoint preserves the same norm. This sort of gives you a natural way of seeing 2 as halfway between 1 and infinity, and giving, for every p, a corresponding q that is equally far away from the middle in the other direction, in the appropriate sense.
Proof of claim: Given p and q such that 1p+1q=1, and a row vector φ=(φ1,...,φn) with Lq norm 1, let xi=|φi|q, so that x1+...+xn=1. Then let vi:=±x1/pi (with the same sign as φi). The column vector v=(v1,...,vn)T has Lp norm 1. φv=φ1v1+...+φnvn=x1p+1q1+...+x1p+1qn=1. This shows that the dual-Lp norm of φ is at least 1. Standard constrained optimization techniques will verify that this v maximizes φv subject to the constraint that v has Lp norm 1, and thus that the dual-Lp norm of φ is exactly 1.
Corollary: If a matrix preserves Lp norm for any p≠2, then it is a permutation matrix (up to flipping the signs of some of its entries).
Proof: Let q be such that 1p+1q=1. The columns of the matrix each have Lp norm 1, so the whole matrix has Lp norm n1/p (since the entries from each of the n columns contribute 1 to the sum). By the same reasoning about its adjoint, the matrix has Lq norm n1/q. Assume wlog p<q. Lq norm is ≤ Lp norm for q>p, with equality only on scalar multiples of basis vectors. So if any column of the matrix isn’t a basis vector (up to sign), then its Lq norm is less than 1; meanwhile, all the columns have Lq norm at most 1, so this would mean that the Lq norm of the whole matrix is strictly less than n1/q, contradicting the argument about its adjoint.