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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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