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Title: Re: eigenvalues property ? Post by MonicaMath on Mar 28th, 2009, 8:29pm maybe you can start with: since r is not an eigenvalue for A then det(r*I - A) not zero, so (A-r*I) is nonsingular, and use the fact that Bx=r*x, then show that || inv(r*I - A) (B-A) ||<1 to obtain a contradiction. |
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Title: Re: eigenvalues property ? Post by trusure on Mar 28th, 2009, 9:33pm I tried your method but there is no result ?! can anyone give me a hint ! ::) |
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Title: Re: eigenvalues property ? Post by Eigenray on Mar 29th, 2009, 5:31pm Well that's a funny problem. Note that (r I - A)-1(B-A) = (r I - A)-1( (r I - A) + (B - r I) ) = I + C, where C = (r I - A)-1(B - r I) is a singular matrix. Do you know the following result: if || I + C || < 1, then C is non-singular? |
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