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Re: Re: Re: Questions regarding MatrixExp, and its usage

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63407] Re: [mg63390] Re: [mg63355] Re: [mg63335] Questions regarding MatrixExp, and its usage
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Fri, 30 Dec 2005 02:32:25 -0500 (EST)
  • References: <200512281024.FAA00331@smc.vnet.net> <50213.70.144.61.64.1135788300.squirrel@webmail.wolfram.com>
  • Sender: owner-wri-mathgroup at wolfram.com

>>>>>
>>>>> In this case, while I am not 100% sure, I tend to believe the
>>>>> situation
>>>>> to be quite analogous. We are interested in the equation
>>>>>
>>>>> MatrixExp[B*p]==MatrixPower[MatrixExp[B],p]
>>>>
>>>>
>>>> Many thanks to Pratik, Daniel, and Andrzej for their very  
>>>> insightful and
>>>> expert feedback!  :)
>>>>
>>>>> I believe this will hold for real matrices B and (probably) all
>>>>> complex
>>>>> p but will not hold in general. In fact I believe most what I   
>>>>> wrote
>>>>> above can be generalised to this case, although the statements   
>>>>> and
>>>>> proofs would be more complicated.
>>>>
>>>>
>>>> Hmm ... actually, from the sample example listed below, I don't  
>>>> believe
>>>> that it will hold *in general* for real B *and* real p:
>>>>
>>>> In[1]: params={theta->Pi^Pi,p->Sqrt[2]};
>>>> In[2]: B=theta {{Cot[theta],Csc[theta]},{-Csc[theta],-Cot[theta]}};
>>>> In[3]: test1=Simplify[MatrixExp[B p]/.params];
>>>> In[4]: test2=Simplify[MatrixPower[MatrixExp[B],p]/.params];
>>>> In[5]: Simplify[test1 == test2]
>>>> Out[5]: False
>>>>
>>>> Daniel has suggested that for (square matrix) B and (scalar) p both
>>>> being
>>>> real-valued, this only will hold if B is positive definite  
>>>> (although I
>>>> suspect that this also may hold with B being positive semi-definite
>>>> too).
>
> What I suggested, or should have in case I misworded it, is that  
> positive
> definiteness is a sufficient assumption for the identity to hold.  
> It may
> hold in other cases as well.
>
> It will hold if
>
> (1) p is integer
>
> or
>
> (2) B (in notation above) is positive definite
>
> (or, I think)
>
> (3) -1<=p<=1.
>
> A case where it perhaps does not hold (I'm trvelling and on the far  
> side
> of the WRI firewall, hence cannot check this):
>
> p = 3/2, B = {{0,-1}, {1,0}} or perhaps the negative thereof. The idea
> being to use a matrix that emulates complex multiplication by Sqrt 
> [-1].

Why, of course! When I wrote "will hold for real matrices B" I was  
just being stupid and thinking of "matrices with real eigenvalues".  
What I really meant was normal matrices with real eigenvalues, which  
I think is probably a sufficient condition, since it seems that in  
this case the condition will hold if it holds for the eigenvalues. Is  
that right?

I was in a great hurry writing all this since out daughter is  
visiting us from Germany for the holidays and we are constantly going  
out or going on trips  and I can only spare a few minutes at a time  
for these messages. But let me quickly explain what I meant when I  
wrote about Jordan canonical forms.  If my memory does not fail me  
there is the following standard procedure for defining f(A) for A is  
any complex matrix where f[x] is any complex  function  for which  
certain derivatives exists. Take the Jordan decomposition of A. Let  
its Jordan canonical form be

A = Inverse[S]. Sum[B_i,{i,1,m}].S

where the B_i are the Jordan blocks. We shall define f[A] as the  
obvious sum once we have defined f[B_i]. So we need only define f 
[B_i]. Suppose the minimal polynomial of B_i is p(x) = (x-r)^k. Then  
we define f[B_i] as the following matrix

Table[KroneckerDelta[i ² j, True] *Derivative[j - i][f][r]/(j - i)!,  
{ i, 1, k}, {j, 1, k}]

The important thing is that all the derivatives required to define  
the above matrices should exist!
If they do f[A] will be well defined.  So, if I am right, taking f 
[x_]:= x^z ought to give the definition of power of a matrix for  
provided all the above derivatives are well defined. If z is integer  
certainly here will be no problems, and the definition will agree  
with the usual one.


Andrzej


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