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Re: Re: Questions regarding MatrixExp, and its usage

  • To: mathgroup at
  • Subject: [mg63366] Re: [mg63355] Re: [mg63335] Questions regarding MatrixExp, and its usage
  • From: Andrzej Kozlowski <akoz at>
  • Date: Tue, 27 Dec 2005 04:42:37 -0500 (EST)
  • References: <>
  • Sender: owner-wri-mathgroup at

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]

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

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