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That's bad news. I got defeated... sad...
What if A is a projection matrix, which only has eigenvalues 0s and 1s?
Thanks a lot!
Robert Israel wrote:
> In article <1142308611.098406.75220@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> <housing2006@xxxxxxxxx> wrote:
>
>
> >If I have a product: C=A*B,
> >
> >where A is a matrix with 0<= eigenvalues <= a, B is a matrix with 0<=
> >eigenvalues <= b,
> >
> >is there a rigorous way for me to prove C=A*B has eigenvalues
> >satisfying 0<=eigenvalues <= a*b ?
>
> It would be true if A and B commute, but it's not true in general.
> For example,
> [ use fixed-width font]
>
> [ 0 0 ]
> A = [ 0 1 ]
>
> [ 2 -2 ]
> B = [ 1 -1 ]
>
> [ 0 0 ]
> AB = [ 1 -1 ]
>
> A and B both have eigenvalues 0 and 1, but AB has eigenvalues 0
> and -1.
>
> Robert Israel israel@xxxxxxxxxxx
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada
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