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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


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