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

**References**:**infinity expression from matrix inverse***From:*"Xiao Huang" <xiaohda@hotmail.com>