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Re: wrong answer or no answer?

  • To: mathgroup at
  • Subject: [mg82812] Re: wrong answer or no answer?
  • From: Roger Bagula <rlbagula at>
  • Date: Wed, 31 Oct 2007 06:19:54 -0500 (EST)
  • References: <> <fg6pf3$ckv$>

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
Andrzej Kozlowski
It certainly does have this property.
Your second post is probably best :

If Det[M]=0, then MatrixPower[M,n]=0 not matter what n is.

SU(2) is sort of the basic Lie Algebra:
I was trying to get a 2by2 matrix weighted graph
of it and see what that resulted in the
M= i-k
type matrix comes out of both graphs.
It doesn't seem to be a productive approach.
It came from the idea that SU(2) was three U(1)'s
and U(1)*SU(2) was, then four U(1).

I want to thank everyone else that posted.
It at least made us all think and
may have taught me some basic matrix theory.
You would expect the U(1)*SU(2) graph to have zero net wight,
but not the SU(2) one.
In theory:  graph is
   A  B
A 0 i+k
B j  0
a=Det[M0]=-i j - j k
Using: ( I used the wrong i,j,k last at night... )
I get :
MatrixPower[a, 1/2]
{{1/2 +I/2, -1/2-I/2}, {1/2 +I/2, 1/2 +I/2}}
(1/2 +I/2)*{{1,-1},{1,1}}
   A           B
A 0         i+k
B j +"1"  0
That gives for U(1)*SU(2):
Clear[i, j, k, O1]
M1 = {{0, i + k}, {j + O1, 0}}
b = Det[M1]
i = {{0, 1}, {1, 0}}
j = {{0, I}, {-I, 0}}
k = {{1, 0}, {0, -1}}
O1 = IdentityMatrix[2]
MatrixPower[b, 1/2]
Det-> I*Sqrt[2]
Maybe I did something wrong again.
This approach to these groups is new to me.
I got it wrong the first time.
Probably wrong the second time too.
Roger Bagula

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