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MathGroup Archive 2004

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Re: infinity expression from matrix inverse

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49810] Re: [mg49800] infinity expression from matrix inverse
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Sun, 1 Aug 2004 18:48:42 -0400 (EDT)
  • References: <200408010810.EAA02530@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Xiao Huang wrote:
> Dear Friends,
> 
> Recently I've been having this problem with Mathematica 5.0: In the follwing 
> program, when I set t=2 to 11, mathematica can give me the inverse of matrix 
> n0, but for t >=12, it fails to produce the inverse due to infinity 
> expressions. Can anyone explain this to me? (I guess the problem occurs in 
> the last line.) Thanks a lot!  Those a1,a2,n0,n0s,nn,c you see in the 
> following program are just some matrices.
> 
> Best Regards,
> 
> Xiao
> 
> \!\(<< LinearAlgebra`MatrixManipulation`\[IndentingNewLine]
>   \(Remove["\<Global`*\>"];\)\[IndentingNewLine]
>   \(ó = 8;\)\[IndentingNewLine]
>   \(t = 11;\)\[IndentingNewLine]
>   \(â = 0.6;\)\[IndentingNewLine]
>   \(a1 = AppendRows[\ \ \ \ IdentityMatrix[\ t - 1\ ]\ , \ \ \ Table[\ 0\ , 
> \ \
> {\ t - 1\ }, \ {\ 1\ }\ ]\ \ \ \ ];\)\[IndentingNewLine]
>   \(a2 = AppendRows[\ \ \ \ Table[\ 0\ , \ {\ t - 1\ }, \ {\ 1\ }\ ]\ , \ \ 
> \
> IdentityMatrix[\ t - 1\ ]\ \ \ \ \ ];\)\[IndentingNewLine]
>   \(nn = Transpose[a1] . a1;\)\[IndentingNewLine]
>   \(n1 = Transpose[a1] . a2;\)\[IndentingNewLine]
>   \(c = IdentityMatrix[t] + AppendColumns[\ Table[0, \ {1}, \ {t}],
>     AppendRows[\ \ \(-â\)*IdentityMatrix[t - 1], \
>   Table[\ 0, \ {t - 1}, \ {1}\ ]\ ]\ ];\)\[IndentingNewLine]
>   \(vc = ó\^2*Inverse[c] . Transpose[Inverse[c]];\)\[IndentingNewLine]
>   \(n0 = IdentityMatrix[t]\  + \ 2  x*vc . nn;\)\[IndentingNewLine]
>   \(Print[\ MatrixForm[n0]\ ];\)\[IndentingNewLine]
>   \(n2 = Inverse[n0] . vc;\)\[IndentingNewLine]
>   \)

I doubt this is specific to version 5. Also, I'd do this using exact 
input, as below (names changed to avoid non-ascii characters).

<<LinearAlgebra`MatrixManipulation`
oo = 8; t = 12; aa = 3/5;
a1 = AppendRows[IdentityMatrix[t-1], Table[0, {t-1}, {1}]];
a2 = AppendRows[Table[0, {t-1}, { 1 } ], IdentityMatrix[t-1]];
nn = Transpose[a1].a1;
n1 = Transpose[a1].a2;
c = IdentityMatrix[t] + AppendColumns[Table[0, {1}, {t}],
    AppendRows[-aa*IdentityMatrix[t - 1], Table[0, {t-1}, {1}]]];
vc = oo^2*Inverse[c].Transpose[Inverse[c]];
n0 = IdentityMatrix[t] + 2*x*vc.nn;

Now we'll do the solving step. It is more efficient to use LinearSolve 
instead of Inverse.

In[57]:=
Timing[n2=LinearSolve[n0,vc,Method\[Rule]OneStepRowReduction];]
Out[57]=
{1.71 Second,Null}

In[41]:=
LeafCount[n2]
Out[41]=
297054

In[58]:=
Timing[n2t=Together[n2];]
Out[58]=
{8.79 Second,Null}

In[59]:=
LeafCount[n2t]
Out[59]=
12970

Your matrix consists of linear polynomials in one variable. As an 
alternative approach that might help for large examples, you might try 
interpolation. For details see

http://groups.google.com/groups?q=Lichtblau+LinearSolve+group:sci.math.symbolic&hl=en&lr=&ie=UTF-8&group=sci.math.symbolic&safe=off&selm=51742f%24iij%40dragonfly.wolfram.com&rnum=3

Moreover, as your matrix has the structure IdentityMatrix+x*M, there 
should also be an efficient power series approach, using knowledge about 
the degree of rational functions in the result.


Daniel Lichtblau
Wolfram Research



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