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wu :: forums    general    wanted (Moderators: ThudnBlunder, towr, Grimbal, SMQ, william wu, Icarus, Eigenray)    Re: eigenvalues property ? « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Re: eigenvalues property ?  (Read 2214 times) MonicaMath Newbie     Gender: Posts: 43 Re: eigenvalues property ?   « on: Mar 28th, 2009, 8:29pm » Quote Modify 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. IP Logged trusure Newbie     Gender: Posts: 24 Re: eigenvalues property ?   « Reply #1 on: Mar 28th, 2009, 9:33pm » Quote Modify I tried your method but there is no result ?!   can anyone give me a hint ! IP Logged Eigenray wu::riddles Moderator Uberpuzzler     Gender: Posts: 1948 Re: eigenvalues property ?   « Reply #2 on: Mar 29th, 2009, 5:31pm » Quote Modify 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? IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis => wanted   - psychology   - chinese « Previous topic | Next topic »

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