Re: Multiply 2 matrices where one contains differential

*To*: mathgroup at smc.vnet.net*Subject*: [mg104463] Re: [mg104417] Multiply 2 matrices where one contains differential*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sun, 1 Nov 2009 04:00:08 -0500 (EST)*References*: <20091029225146.K51SM.569239.imail@eastrmwml30>*Reply-to*: drmajorbob at yahoo.com

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