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Re: wrong answer or no answer?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82720] Re: [mg82710] wrong answer or no answer?
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 30 Oct 2007 03:18:23 -0500 (EST)
  • References: <200710291054.FAA06915@smc.vnet.net> <06386FCE-05E8-406E-AA04-59745F34DA53@mimuw.edu.pl>

I forgot to note that MatrixPower seems to return the zero matrix for  
any fractional power of a nilpotent matrix. This can be seen in the  
case of a strict upper triangular matrix (every nilpotent matrix is  
similar either to a strictly upper or strictly lower triangular matrix):

AA = Normal[SparseArray[{{i_, j_} /; i < j :>
            a[i, j]}, {5, 5}]]
  {{0, a[1, 2], a[1, 3], a[1, 4], a[1, 5]},
    {0, 0, a[2, 3], a[2, 4], a[2, 5]},
    {0, 0, 0, a[3, 4], a[3, 5]}, {0, 0, 0, 0, a[4, 5]},
    {0, 0, 0, 0, 0}}

  MatrixPower[AA, 1/2]
( {
   {0, 0, 0, 0, 0},
   {0, 0, 0, 0, 0},
   {0, 0, 0, 0, 0},
   {0, 0, 0, 0, 0},
   {0, 0, 0, 0, 0}
  } )

This probably is a bug, since (as I wrote below) it is trivial to  
prove that such powers do not exist (and any positive power of the  
zero matrix is the zero matrix).


Andrzej Kozlowski

On 29 Oct 2007, at 21:52, Andrzej Kozlowski wrote:

>
> Your matrix M has the property:
>
> MatrixPower[M, 2]
> {{0, 0}, {0, 0}}
>
> It is easy to prove using elementary linear algebra that such a  
> matrix has no square root. In fact one can prove more. Suppose than  
> M is an n by n matrix such that M^n=0 but M^(n-1) !=0 (in other  
> words M is nilpotent of order n). Then M has no square root.
>
> The proof is easy so I won't bother to give it here.
>
> Andrzej Kozlowski
>
>
> On 29 Oct 2007, at 19:54, Roger Bagula 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
>>
>



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