Re: Multiply 2 matrices where one contains differential

*To*: mathgroup at smc.vnet.net*Subject*: [mg104468] Re: [mg104417] Multiply 2 matrices where one contains differential*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Sun, 1 Nov 2009 04:01:05 -0500 (EST)*References*: <20091029225146.K51SM.569239.imail@eastrmwml30>*Reply-to*: drmajorbob at yahoo.com

You could also simply multiply A*B, if A and B are the same dimensions. Anyway... I gave a solution. If somebody has a better one, we'll see. Bobby On Sat, 31 Oct 2009 16:52:04 -0500, Nasser M. Abbasi <nma at 12000.org> wrote: > 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: 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 > -- DrMajorBob at yahoo.com

**Re: Multiply 2 matrices where one contains differential**

**Re: Multiply 2 matrices where one contains differential**

**Re: Multiply 2 matrices where one contains differential**

**Re: Multiply 2 matrices where one contains differential**