Re: wrong answer or no answer?
- To: mathgroup at smc.vnet.net
- Subject: [mg82812] Re: wrong answer or no answer?
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Wed, 31 Oct 2007 06:19:54 -0500 (EST)
- References: <200710291054.FAA06915@smc.vnet.net> <fg6pf3$ckv$1@smc.vnet.net>
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 M0={{0,i+k},{j,0}} a=Det[M0]=-i j - j k Using: ( I used the wrong i,j,k last at night... ) i={{0,1},{1,0}} j={{0,I},{-I,0}} k={{1,0},{0,-1}} I get : a={{0,-I},{I,0}} MatrixPower[a, 1/2] {{1/2 +I/2, -1/2-I/2}, {1/2 +I/2, 1/2 +I/2}} or (1/2 +I/2)*{{1,-1},{1,1}} Det->I Trying: 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
- References:
- wrong answer or no answer?
- From: Roger Bagula <rlbagula@sbcglobal.net>
- wrong answer or no answer?