Re: wrong answer or no answer?

*To*: mathgroup at smc.vnet.net*Subject*: [mg82740] Re: wrong answer or no answer?*From*: Peter Pein <petsie at dordos.net>*Date*: Tue, 30 Oct 2007 03:28:45 -0500 (EST)*References*: <fg4e5h$6oa$1@smc.vnet.net>

Roger Bagula schrieb: > 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 > Hi Roger, MatrixPower[{{-1,I},{I,1}},n] gives a matrix with KroneckerDeltas in it. I guess, MatrixPower is not well defined for non-integral n. Try to approximate KroneckerDelta with Exp[-k*x²] (this is Version 5.2): In[1]:= p0 = MatrixPower[{{-1, I}, {I, 1}}, n] /. KroneckerDelta -> (Exp[(-k)*#1^2] & ) /. n -> 1/2 Out[1]= {{0, I/E^(k/4)}, {I/E^(k/4), 2/E^(k/4)}} obviously, the limit for k->oo is the nullmatrix. But let's go on with this expression and solve for the elements independently (by mapping Reduce): In[2]:= (Reduce[#1 == 0, k] & ) /@ Flatten[p0 . p0 - {{-1, I}, {I, 1}}] Out[2]= {C[1] \[Element] Integers && k == 4*I*Pi*C[1], C[1] \[Element] Integers && k == 2*(2*I*Pi*C[1] + Log[2]), C[1] \[Element] Integers && k == 2*(2*I*Pi*C[1] + Log[2]), C[1] \[Element] Integers && k == 2*(2*I*Pi*C[1] + Log[3])} finally allow different C[i] for each equation and try to solve simultaneously: In[3]:= Reduce[MapIndexed[#1 /. C[1] -> C @@ #2 & , %], k, Backsubstitution -> True] Out[3]= (C[4] | C[3] | C[2] | C[1]) \[Element] Integers && C[3] == C[4] + (I*(Log[2] - Log[3]))/(2*Pi) && C[2] == C[4] + (I*(Log[2] - Log[3]))/(2*Pi) && C[1] == C[4] - (I*Log[3])/(2*Pi) && k == 4*I*Pi*C[4] + 2*Log[3] To take only the simple third equation: C[1] == C[4] - (I*Log[3])/(2*Pi) means that Log[3] is an integral multiple of 2*Pi*I, which seems funny to me. ;-) Chances are really bad to find a matrix m with m.m=={{-1,I},{I,1}}: In[4]:= GroebnerBasis[Flatten[(#1 . #1 & )[mat = Array[a, {2, 2}]] - {{-1, I}, {I, 1}}], Flatten[mat]] Out[4]= {1} Regards, Peter