Re: wrong answer or no answer?

*To*: mathgroup at smc.vnet.net*Subject*: [mg82796] Re: [mg82710] wrong answer or no answer?*From*: DrMajorBob <drmajorbob at bigfoot.com>*Date*: Wed, 31 Oct 2007 06:11:40 -0500 (EST)*References*: <21140268.1193712179571.JavaMail.root@m35>*Reply-to*: drmajorbob at bigfoot.com

No solution exists. Here's a demonstration that's pretty convincing, I think: m2={{-1,x},{x,1}}; r={{a,b},{c,d}}; r.r-m2//Flatten; Thread[%==0]; soln=Solve[%,{a,b,c,d}]//Simplify r.r==m2/.soln//Simplify soln/.x->I//Quiet {{a->-((1+x^2+Sqrt[-1-x^2])/(Sqrt[2] (-1-x^2)^(3/4))),d->(-1-x^2+Sqrt[-1-x^2])/(Sqrt[2] (-1-x^2)^(3/4)),b->x/(Sqrt[2] (-1-x^2)^(1/4)),c->x/(Sqrt[2] (-1-x^2)^(1/4))},{a->-((\[ImaginaryI] (-1-x^2+Sqrt[-1-x^2]))/(Sqrt[2] (-1-x^2)^(3/4))),d->(\[ImaginaryI] (1+x^2+Sqrt[-1-x^2]))/(Sqrt[2] (-1-x^2)^(3/4)),b->(\[ImaginaryI] x)/(Sqrt[2] (-1-x^2)^(1/4)),c->(\[ImaginaryI] x)/(Sqrt[2] (-1-x^2)^(1/4))},{a->(\[ImaginaryI] (-1-x^2+Sqrt[-1-x^2]))/(Sqrt[2] (-1-x^2)^(3/4)),d->-((\[ImaginaryI] (1+x^2+Sqrt[-1-x^2]))/(Sqrt[2] (-1-x^2)^(3/4))),b->-((\[ImaginaryI] x)/(Sqrt[2] (-1-x^2)^(1/4))),c->-((\[ImaginaryI] x)/(Sqrt[2] (-1-x^2)^(1/4)))},{a->(1+x^2+Sqrt[-1-x^2])/(Sqrt[2] (-1-x^2)^(3/4)),d->(1+x^2-Sqrt[-1-x^2])/(Sqrt[2] (-1-x^2)^(3/4)),b->-(x/(Sqrt[2] (-1-x^2)^(1/4))),c->-(x/(Sqrt[2] (-1-x^2)^(1/4)))}} {True,True,True,True} {{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity},{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity},{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity},{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity}} Visually examining the poles in Solve's output (as x tends to I), one can see that the Indeterminate results are actually infinite, like the others. Bobby On Mon, 29 Oct 2007 05:54:37 -0500, Roger Bagula <rlbagula at sbcglobal.net> wrote: > M = {{-1, I}, {I, 1}} > MatrixPower[M, 1/2] > gives > {{0, 0}, {0, 0}} > > So try it as {{a,b},{c,d}} squared: > c = b; a = I*Sqrt[1 + b^2]; d = Sqrt[1 - b^2 > FullSimplify[{a^2 + b c + 1 == 0, a b + b d - I == 0, a c + > c d - I == 0, b c + d^2 - 1 == 0}] > {True, -I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 + > b^2)]== 0, -I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 + > b^2)]== 0, True} > Solve[-I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 + > b^2)]== 0,b] > {} > which says there is no solution. > > This problem comes from the graph of SU(2) and of U(1)*SU(2) > as a two vertex 3 directed connections (i,j,k) and a 2 vertex with 4 > directed connections (1,i,j,k}. > Basically there is either a solution or there is none. > Mathematica gives zero and the null set from two different approaches. > {{0,i+k},{j,0}} and {{0,i+k},{j+IdentityMatrix[2],0}} > > It is pretty much a break down of mathematical definitions. > The matrix M does appear to have not one, but four solutions > the way I do it: > > M2={{+/-I*Sqrt[1 + b^2], b}, {b, +/-Sqrt[1 + b^2]}} > I really may be doing it all wrong. > b=+/-Sqrt[+/-1/2+I/2] > which gives the stange answers from this code: > Clear[b] > M2 = {{I*Sqrt[1 + b^2], b}, {b, Sqrt[1 + b^2]}} > Det[M2] > Solve[Det[M2] == 0, b] > b0 = b /. Solve[Det[M2] == 0, b][[2]] > M20 = {{-I*Sqrt[1 + b0^2], b0}, {b0, Sqrt[1 + b0^2]}} > FullSimplify[M20] > FullSimplify[M20.M20] > > All this leaves me really puzzled. > Usually Mathematica takes away my doubts, > but here it isn't any help at all. > Maybe it is a paradox? > Roger Bagula > > -- DrMajorBob at bigfoot.com