Re: give the diagonal of m?

*To*: mathgroup at smc.vnet.net*Subject*: [mg128580] Re: give the diagonal of m?*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Wed, 7 Nov 2012 00:56:35 -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>

There are more straightforward ways to get a diagonal A = Array[a, {4, 4}]; Diagonal[A] {a[1, 1], a[2, 2], a[3, 3], a[4, 4]} List @@ Tr[A] (* works only with symbolic matrix *) {a[1, 1], a[2, 2], a[3, 3], a[4, 4]} Tr[A, List] {a[1, 1], a[2, 2], a[3, 3], a[4, 4]} Transpose[A, {1, 1}] {a[1, 1], a[2, 2], a[3, 3], a[4, 4]} % === %% === %%% === %%%% True The diagonal does not have to be the main diagonal Diagonal[A, #] & /@ Range[-3, 3] {{a[4, 1]}, {a[3, 1], a[4, 2]}, {a[2, 1], a[3, 2], a[4, 3]}, {a[1, 1], a[2, 2], a[3, 3], a[4, 4]}, {a[1, 2], a[2, 3], a[3, 4]}, {a[1, 3], a[2, 4]}, {a[1, 4]}} B = Partition[Range[16], 4]; Diagonal[B] {1, 6, 11, 16} List @@ Tr[B] (* does not work with numeric matrix *) 34 Tr[B, List] {1, 6, 11, 16} Transpose[B, {1, 1}] {1, 6, 11, 16} % === %% === %%% === %%%% False %% === %%% === %%%%% True Diagonal[B, #] & /@ Range[-3, 3] {{13}, {9, 14}, {5, 10, 15}, {1, 6, 11, 16}, {2, 7, 12}, {3, 8}, {4}} Bob Hanlon On Mon, Nov 5, 2012 at 6:39 PM, <marty.musatov at gmail.com> wrote: > 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 >