Re: give the diagonal of m?

*To*: mathgroup at smc.vnet.net*Subject*: [mg128573] Re: give the diagonal of m?*From*: marty.musatov at gmail.com*Date*: Mon, 5 Nov 2012 18:39:01 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <DD6GKI.3sG@wri.com>

On Saturday, August 12, 1995 12:00:00 AM UTC-7, Richard Mercer wrote: > > An elegant test. My question: why does Transpose[m, {1, > > 1}] return the list of diagonal elements? Yes. Looking under > > Transpose in Wolfram's Mathematica Reference Guide (an > > appendix to Mathematica, 2nd ed.), I find: > > > Transpose[list, {n1,n2, ...}] transposes list so that the kth level > > in list is the nkth level in the result. > > > I expected Depth[Transpose[list, {n1,n2, ...}]] to be > > the same as Depth[list]. I expected Transpose to swap > > levels; I did not expect it to throw any away. Evidently > > I am laboring under some misconception. Can someone > > explain what is going on? Yes. > > I think so, at the price of a certain amount of abstraction and background > discussion. > > Consider the matrix > > A = {{16,27},{38,49}}, or > A = [ 16 27 ] > [ 38 49 ] > Then > C = Transpose[A,{1,1}] gives > {16,49} > Why? Yes. > > *Short Version* > Since {1,1} is not a permutation, Transpose[list,{1,1}] is not defined by the > Reference Guide. The effect of using {1,1} is to say the two entries must be > the same, so C[[1]] = A[[1,1]] = 16, and C[[2]] = A[[2,2]] = 49. The effect is > to compress a two-dimensional list into a one-dimensional list. > > If you believe that, I have some Netscape shares I'd like to sell you! > So you'd better read the... > > > *Long Version* > (0) The discussion below works just as well for non-square matrices and their > higher-dimensional analogues, but it's simplest to consider a 2x2 matrix. > > (1) The structured list objects on which Transpose operates can be viewed as > functions on certain integer lattices. Sounds fancy, but an example shows it's > really a simple idea. > > The matrix A can be thought of as a function of two indices with the following > values: > A(1,1) = 16 > A(1,2) = 27 > A(2,1) = 38 > A(2,2) = 49 > (In Mathematica you use double brackets instead of parentheses.) > The domain of the function A is therefore the set of ordered pairs > D = {(1,1), (1,2), (2,1), (2,2)}. > (set notation, not Mathematica notation!) > D is the same as the set (cartesian) product {1,2} X {1,2}. > > (2) The ordinary transpose operation can be viewed as the composition of the > function A and the map t defined from D to D as follows: > t(1,1) = (1,1) > t(1,2) = (2,1) > t(2,1) = (1,2) > t(2,2) = (2,2) > > So if B = Transpose[A], then > B(1,1) = A(t(1,1)) = A(1,1) = 16 > B(1,2) = A(t(1,2)) = A(2,1) = 38 > B(2,1) = B(t(2,1)) = A(1,2) = 27 > B(2,2) = B(t(2,2)) = A(2,2) = 49 > Therefore B = {{16,38},{27,49}} in Mathematica notation, or > B = [ 16 38 ] > [ 27 49 ] > as expected. > > (3) The key is to understand the map t and how it can be generalized to other > situations. The map t is defined by a "permutation" or reordering applied to > the elements of the domain D. In this case the reordering is described by {2,1} > because > > -- the first entry of the result is the second(2) entry of the input, > -- the second entry of the result is the first(1) entry of the input, i.e. > t(a,b) = (b,a). Therefore > Transpose[A,{2,1}] means the same as Transpose[A], and > Transpose[A,{1,2}] = A, > because the permutation {1,2} means no reordering at all. > > (4) If we now extend this discussion to the non-permutation {1,1}, it should > define a map s by > -- the first entry of the result is the first(1) entry of the input, > -- the second entry of the result is the first(1) entry of the input, > > i.e. s(a) = (a,a). > Why not s(a,b) = (a,a)? Yes. Since only the first entry of the input is used, we > assume the input has only one entry. > s maps from {1,2} into {1,2} X {1,2} = D. > > (5) By analogy, C = Transpose[A,{1,1}] should now represent the composition of > the function A with the map s. The result is a function on {1,2}, with values > C(1) = A(s(1)) = A(1,1) = 16 > C(2) = A(s(2)) = A(2,2) = 49 > > This discussion could be continued to cases of higher dimensional lists, but > I'm amazed if you've read this far! > > Exercise for the reader: > If you have a three dimension list, e.g. > A = {{{16,27},{38,49}},{{91,82},{73,64}}} > what should be the result of Transpose[A,{1,1,2}] ? Yes. > Explain why Mathematica accepts the non-permutation {1,1,2} but not {1,1,3} in > its place. > > Richard Mercer AN ... integers for all j = 1, ... , n, xj = 0, 1; and the rank of the matrix (aj) equals 2.Musatov

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