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February 2nd, 2010, 05:26 PM   #1
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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 counter-example 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.
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February 3rd, 2010, 08:50 AM   #2
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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/
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February 7th, 2010, 11:57 AM   #3
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Re: prove it's an eigenvalue

Quote:
Originally Posted by pascallimacon
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.

tnaks, that is really a great help, now i am clear.

http://ficus.pntic.mec.es/~frej0002/
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