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 April 6th, 2010, 12:09 PM #1 Senior Member   Joined: Nov 2009 Posts: 169 Thanks: 0 Find an invertible matrix P Let $T:R^4 --> R^4$be a linear transformation T(v) = Av where A = [1 1 3 1] [0 2 2 4] [0 0 3 2] [0 0 1 4] Find the characteristic polynomial of A and compute the eigenvalues of T. Find a basis for each eigenspace. Either find an invertible matrix P and a diagonal matrix D such that $P^{-1}AP= D$or else explain why this is impossible. I know how to find the characteristic polynomial and the eigenvalues of A but not T.
 June 24th, 2010, 07:13 AM #2 Member   Joined: Jun 2010 Posts: 80 Thanks: 0 Re: Find an invertible matrix P Hello, nice to meet you. I'm new in here. I think your problem resolves to : The eigenvalues and carac. polynomial of A are of T. This, because it can be proven that any representation of T has the same eigenvalues and carac. polynomial : for example : Ax=kx, then I write x=Py a change of basis, and multiply leftly with P^-1 : P^-1APy=ky, where we recognize that P^-1AP=A' is the matrix representing A in the new basis, hence the result. So we can define these quantities intrinsically, in the sense : even if the calculation is made in a basis, the basis has no influence, since a change of basis does not change these results.

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