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