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April 6th, 2010, 12:09 PM   #1
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Find an invertible matrix P

Let 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 or else explain why this is impossible.



I know how to find the characteristic polynomial and the eigenvalues of A but not T.
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June 24th, 2010, 07:13 AM   #2
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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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