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 April 3rd, 2010, 03:58 PM #1 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 eigenvalue Let T be a linear operator on a finite-dimensional vector space V and let B be an ordered basis for V. Prove that lama(y) is an eigenvalue of T if and only if y is an eigenvalue of [T]_B How do I start. I know you need to use det(A-yI) to find the eigenvalue.
 April 4th, 2010, 02:53 PM #2 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: eigenvalue Do you know the spectral theorem? Alternately, do you know the completeness relation and use outer products to express diagonalizeable matrices?
 April 4th, 2010, 11:09 PM #3 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 Re: eigenvalue no sorry. can we use characteristic polynomial? I have T(v)=lambda v or (T-lambda Inxn)=0.
 April 5th, 2010, 12:56 AM #4 Senior Member   Joined: Apr 2008 Posts: 435 Thanks: 0 Re: eigenvalue I'm afraid to say that I don't know what machinery you have to work with. But intuitively, this must be true. Perhaps I'll go through my intuition, and you can think of how to prove that. If lambda is an eigenvalue of T, then there is some vector v such that Tv = (lambda)v, and so when you stick in a particular vector into this transformation, you get a lambda multiple of the vector out. Now we change bases. Now, the new vector will still go to the lambda multiple of the vector v written in the new basis, as we haven't changed the transformation or the characteristics of the space. Does that help?

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