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Indeed you are right. The IVT will give a matrix such that real part or
the imaginary part of the trace is zero, but not necessarily both at the
same time. I realized this yesterday, but the "cancel post" function
doesn't seem to work too well...
--
Daniel Mayost
In article <1162311247.286120.241490@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Robert Israel <israel@xxxxxxxxxxx> wrote:
>
>Daniel Mayost wrote:
>> I don't know if this helps, but there exist unitary matrices Q,R,S,T,U
>> such that:
>>
>> Tr(AQ)=Tr(BR)=Tr(CS)=Tr(DT)=Tr(EU)=0
>>
>> Take A as an example. Note that both I and -I are unitary. Since
>> the space of unitary matrices is connected, there exists a continuous
>> path from I to -I in this space. Since Tr(AI)=Tr(A) and Tr(-AI)=-Tr(A),
>> by the Intermediate Value Theorem there exists a unitary matrix U along
>> this path such that Tr(AU)=0.
>
>There do exist such unitary matrices (for n x n matrices if n > 1, as I
>
>remarked in my other posting in this thread), but the Intermediate
>Value Theorem does not prove it, because it doesn't apply
>to complex-valued functions. Note that the result would not be true
>for n=1, even though U(1) is connected.
>
>Robert Israel israel@xxxxxxxxxxx
>Department of Mathematics http://www.math.ubc.ca/~israel
>University of British Columbia Vancouver, BC, Canada
>
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