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Re: Re: Questions regarding MatrixExp, and its usage
*To*: mathgroup at smc.vnet.net
*Subject*: [mg63366] Re: [mg63355] Re: [mg63335] Questions regarding MatrixExp, and its usage
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Tue, 27 Dec 2005 04:42:37 -0500 (EST)
*References*: <200512250719.CAA01655@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
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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