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MathGroup Archive 2005

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Re: Re: Questions regarding MatrixExp, and its usage

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63380] Re: [mg63355] Re: [mg63335] Questions regarding MatrixExp, and its usage
  • From: "Michael Chang" <michael_chang86 at hotmail.com>
  • Date: Wed, 28 Dec 2005 03:55:37 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

>From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
To: mathgroup at smc.vnet.net
>To: Michael Chang <michael_chang86 at hotmail.com>
>Subject: [mg63380] Re: [mg63355] Re: [mg63335] Questions regarding MatrixExp, and its 
>usage
>Date: Mon, 26 Dec 2005 23:47:09 +0900
>
>*This message was transferred with a trial version of CommuniGate(tm) Pro*
>
>On 25 Dec 2005, at 16:19, Michael Chang wrote:
>
>>>>>>I was therefore wondering if
>>>>>>
>>>>>>MatrixExp[A p]==(MatrixExp[A]^p)
>>>>>>
>>>>>>where 'p' is an arbitrary complex number, and the '^' operator  is my
>>>>>>attempt to denote the matrix power, and *not* an element-by- element
>>>>>>power for each individual matrix entry.  Or does such an  expression
>>>>>>only hold for real-valued square A matrices?  Or am I  completely lost
>>>>>>here ...?
>
>
>This can't possibly be true for arbitrary square complex matrices  since it 
>is not even true for matrices of dimension 1, that is  complex numbers.
>In other words, it is not true that Exp[a p]== Exp[a]^p, were a and p  are 
>arbitrary complex numbers. In fact,   Mathematica alone can find  for you 
>an example where this is not true. To see that let's define a  function f 
>of two variables:
>
>
>f[a_, b_] := Exp[a*b] - Exp[a]^b
>
>
>If the identity you held for all complex numbers f would have to be  
>identically zero. However, we can get Mathematica to find an example  when 
>it is not:
>
>
>FindInstance[f[a, b] != 0, {a, b}]
>
>
>{{a -> -(47/10) + (181*I)/10,
>    b -> 91/10 + (122*I)/5}}
>
>Since this seems a little  hard to verify without Mathematica and  since 
>mathematica is sometimes wrong ;-) it may be more convincing to  construct 
>an example by hand. In fact it is pretty easy. All you need  is the well 
>known identity:
>
>Exp[I *A] = Cos[A]+ I *Sin[A]

Sorry for my tardy response ... I'm trying to catch up on my email!  Just a 
quick typo correction ... I believe that the first "A" definition below 
should be A=-Pi, given the definition of Exp[I*A] above ...

Regardless, the key point that Andrzej has made for complex-valued (square) 
matrices is duly noted!

>Put A = -I*Pi. We get Exp[-I*Pi] = -1
>
>Put A = -Pi/2. We get Exp[-Pi/2*I]= -I
>
>But now note that:
>
>Exp[-Pi/2*I] == Exp[-Pi*I *(1/2)]
>
>so if the identity is true than  -I == Exp[-Pi/2*I] == Exp[-Pi*I]^ (1/2) = 
>(-1)^(1/2) == I.
>
>Thus we get a contradiction.
>
>Andrzej Kozlowski

Regards,

Michael



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