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February 2nd, 2010, 06:26 PM  #1 
Senior Member Joined: Nov 2009 Posts: 169 Thanks: 0  prove it's an eigenvalue
Soppose that are linear transformations, if and are eigenvalues for T and S respectively, either prove that is an eigenvalue for ToS or else give a counterexample to show that this is false. Is an eigenvalue for ToT? (Justify!) I am really not familliar with eigenvalue, I know how to find the eigenvalue, but not proving it. 
February 3rd, 2010, 09:50 AM  #2 
Newbie Joined: Dec 2007 From: Madrid Posts: 7 Thanks: 0  Re: prove it's an eigenvalue
Consider defined by: Then, The only eigenvalue of is , the only of is . The eigenvalues of are and , so is not eigenvalue of . On the other hand, if is an eigenvalue of there exists with such that . Then, This proves that is also an eigenvalue of Regards. http://ficus.pntic.mec.es/~frej0002/ 
February 7th, 2010, 12:57 PM  #3  
Senior Member Joined: Nov 2009 Posts: 169 Thanks: 0  Re: prove it's an eigenvalue Quote:
 

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