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*: [mg104466] Re: [mg104417] Multiply 2 matrices where one contains differential operators with one that contains functions of x and y*From*: "Nasser M. Abbasi" <nma at 12000.org>*Date*: Sun, 1 Nov 2009 04:00:42 -0500 (EST)*References*: <20091029225146.K51SM.569239.imail@eastrmwml30> <200910310650.BAA13104@smc.vnet.net> <op.u2on0gpdtgfoz2@bobbys-imac.local>*Reply-to*: "Nasser M. Abbasi" <nma at 12000.org>

DrMajorBob; Thanks for your detailed reply, I need to look at your code more closely. But just to answer your initial point > That doesn't look like a Dot product at all, since it operates a 3x2 > matrix on a 2x3 and gets a 3x3. Yes. I am doing matrix products ofcourse, so 3x2 matrix when multiplied by 2x3 matrix better result in 3x3 or else something is really wrong :) >From help on Dot "The standard (built-in) usage still exists a.b.c or Dot[a, b, c] gives products of vectors, matrices and tensors" I use A.B to multiply matrices by each others? may be I am missing your point here. How else would you multiply matrices with each others in Mathematica other than using the Dot? Best, --Nasser ----- Original Message ----- Subject: [mg104466] Re: [mg104417] Multiply 2 matrices where one contains differential operators with one that contains functions of x and y > That doesn't look like a Dot product at all, since it operates a 3x2 > matrix on a 2x3 and gets a 3x3. > > So... to find out what's really happening: > > a = Array[f, {3, 2}] > b = Array[g, {2, 3}] > {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]]}]]] > r > Dimensions@r > > {{f[1, 1], f[1, 2]}, {f[2, 1], f[2, 2]}, {f[3, 1], f[3, 2]}} > > {{g[1, 1], g[1, 2], g[1, 3]}, {g[2, 1], g[2, 2], g[2, 3]}} > > {{{f[1, 1][g[1, 1]] + f[1, 2][g[2, 1]]}, {f[1, 1][g[1, 2]] + > f[1, 2][g[2, 2]]}, {f[1, 1][g[1, 3]] + > f[1, 2][g[2, 3]]}}, {{f[2, 1][g[1, 1]] + > f[2, 2][g[2, 1]]}, {f[2, 1][g[1, 2]] + > f[2, 2][g[2, 2]]}, {f[2, 1][g[1, 3]] + > f[2, 2][g[2, 3]]}}, {{f[3, 1][g[1, 1]] + > f[3, 2][g[2, 1]]}, {f[3, 1][g[1, 2]] + > f[3, 2][g[2, 2]]}, {f[3, 1][g[1, 3]] + f[3, 2][g[2, 3]]}}} > > {3, 3, 1} > > That looks like an outer product of the rows of a with the columns of b. > > Here's a somewhat similar structure (with one less low-level dimension): > > Outer[h, a, Transpose@b, 1, 1] > Dimensions@% > > {{h[{f[1, 1], f[1, 2]}, {g[1, 1], g[2, 1]}], > h[{f[1, 1], f[1, 2]}, {g[1, 2], g[2, 2]}], > h[{f[1, 1], f[1, 2]}, {g[1, 3], g[2, 3]}]}, {h[{f[2, 1], > f[2, 2]}, {g[1, 1], g[2, 1]}], > h[{f[2, 1], f[2, 2]}, {g[1, 2], g[2, 2]}], > h[{f[2, 1], f[2, 2]}, {g[1, 3], g[2, 3]}]}, {h[{f[3, 1], > f[3, 2]}, {g[1, 1], g[2, 1]}], > h[{f[3, 1], f[3, 2]}, {g[1, 2], g[2, 2]}], > h[{f[3, 1], f[3, 2]}, {g[1, 3], g[2, 3]}]}} > > {3, 3} > > It remains to figure out what "h" should be. A simple solution is > > h[{a_, b_}, {c_, d_}] := a[c] + b[d] > Outer[h, a, Transpose@b, 1, 1] > Dimensions@% > > {{f[1, 1][g[1, 1]] + f[1, 2][g[2, 1]], > f[1, 1][g[1, 2]] + f[1, 2][g[2, 2]], > f[1, 1][g[1, 3]] + f[1, 2][g[2, 3]]}, {f[2, 1][g[1, 1]] + > f[2, 2][g[2, 1]], f[2, 1][g[1, 2]] + f[2, 2][g[2, 2]], > f[2, 1][g[1, 3]] + f[2, 2][g[2, 3]]}, {f[3, 1][g[1, 1]] + > f[3, 2][g[2, 1]], f[3, 1][g[1, 2]] + f[3, 2][g[2, 2]], > f[3, 1][g[1, 3]] + f[3, 2][g[2, 3]]}} > > {3, 3} > > To get back the low-level (probably unnecessary) dimension: > > h[{a_, b_}, {c_, d_}] := {a[c] + b[d]} > Outer[h, a, Transpose@b, 1, 1] > Dimensions@% > > {{{f[1, 1][g[1, 1]] + f[1, 2][g[2, 1]]}, {f[1, 1][g[1, 2]] + > f[1, 2][g[2, 2]]}, {f[1, 1][g[1, 3]] + > f[1, 2][g[2, 3]]}}, {{f[2, 1][g[1, 1]] + > f[2, 2][g[2, 1]]}, {f[2, 1][g[1, 2]] + > f[2, 2][g[2, 2]]}, {f[2, 1][g[1, 3]] + > f[2, 2][g[2, 3]]}}, {{f[3, 1][g[1, 1]] + > f[3, 2][g[2, 1]]}, {f[3, 1][g[1, 2]] + > f[3, 2][g[2, 2]]}, {f[3, 1][g[1, 3]] + f[3, 2][g[2, 3]]}}} > > {3, 3, 1} > > For your actual problem, that's > > 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}}; > h[{a_, b_}, {c_, d_}] := {a[c] + b[d]} > Outer[h, a, Transpose@b, 1, 1] > > {{{y}, {3 x^2 y}, {3}}, {{0}, {x^4}, {2 y}}, {{2 + x}, {x^3 + > 4 x^3 y}, {2 y}}} > > which agrees with your nested For method. > > The function h could be done differently, since it's analogous to > > Inner[Compose, {aa, bb}, {c, d}, Plus] > > aa[c] + bb[d] > > For instance: > > 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}}; > h = Inner[Compose, ##, Plus] &; > Outer[h, a, Transpose@b, 1, 1] > > {{y, 3 x^2 y, 3}, {0, x^4, 2 y}, {2 + x, x^3 + 4 x^3 y, 2 y}} > > That omits the low-level List, but it can be added in a hundred different > ways. For instance, > > 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}}; > h = Inner[Compose, ##, Plus] &; > Map[List, Outer[h, a, Transpose@b, 1, 1], {2}] > > {{{y}, {3 x^2 y}, {3}}, {{0}, {x^4}, {2 y}}, {{2 + x}, {x^3 + > 4 x^3 y}, {2 y}}} > > Bobby > > On Sat, 31 Oct 2009 01:50:39 -0500, Nasser M. Abbasi <nma at 12000.org> > 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 >> >> > > > -- > DrMajorBob at yahoo.com