Re: Multiply 2 matrices where one contains differential operators with one that contains functions of x and y

*To*: mathgroup at smc.vnet.net*Subject*: [mg104457] Re: [mg104417] Multiply 2 matrices where one contains differential operators with one that contains functions of x and y*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 1 Nov 2009 03:58:58 -0500 (EST)*References*: <20091029225146.K51SM.569239.imail@eastrmwml30> <200910310650.BAA13104@smc.vnet.net>

On 31 Oct 2009, at 15:50, Nasser M. Abbasi wrote: > Hello, > Version 7 > > Lets say A is a 3 by 2 matrix, which contains differential operators > in some > entries and 0 in all other entries, as in > > A= { { d/dx , 0 } , {0 , d/dy } , { d/dy , d/dx } } > > And I want to multiply the above with say a 2 by 3 matrix whose > entries are > functions of x and y as in > > B = {{x*y, x^3*y, 3*x + y^2}, {2*x, x^4*y, y^2}} > > I'd like to somehow be able to do A.B, but ofcourse here I can't, as > I need > to "apply" the operator on each function as the matrix > multiplication is > being carried out. > > I tried to somehow integrate applying the operators in A into the > matrix > multiplication of A by B, but could not find a short "functional" way. > > So I ended up solving this by doing the matrix multiplication by > hand using > for loops (oh no) so that I can 'get inside' the loop and be able to > apply > the operator to each entry. This is my solution: > > > A = {{D[#1, x] & , 0 & }, {0 & , D[#1, y] & }, {D[#1, y] & , D[#1, > x] & }} > B = {{x*y, x^3*y, 3*x + y^2}, {2*x, x^4*y, y^2}} > > {rowsA, colsA} = Dimensions[A]; > {rowsB, colsB} = Dimensions[B]; > > r = Table[0, {rowsA}, {colsB}]; (*where the result of A.B goes *) > > For[i = 1, i <= rowsA, i++, > For[j = 1, j <= colsB, j++, > For[ii = 1, ii <= rowsB, ii++, > r[[i,j]] = r[[i,j]] + A[[i,ii]] /@ {B[[ii,j]]} > ] > ] > ] > > MatrixForm[r] > > The above work, but I am sure a Mathematica expert here can come up > with a > true functional solution or by using some other Mathematica function > which I > overlooked to do the above in a more elegent way. > > --Nasser > > Probably the easiest way is to use Inner, which is a generalisation of Dot, replacing Times with the obsolete but still sometimes useful function Compose: Inner[Compose, {{D[#1, x] & , 0 & }, {0 & , D[#1, y] & }, {D[#1, y] & , D[#1, x] & }}, {{x*y, x^3*y, 3*x + y^2}, {2*x, x^4*y, y^2}}, Plus] {{y, 3*x^2*y, 3}, {0, x^4, 2*y}, {x + 2, 4*y*x^3 + x^3, 2*y}} If you do not like to use a no-longer documented function you can replace it by #1[#2]&. Andrzej Kozlowski