The eigenvalues of are exactly the eigenvalues of . Indeed, if then .
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 disgonal 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.