TomMe
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Suppose A: V -> V a linear transformation, with dim(V) = n. If the characteristic polynomial f_{A} of this matrix can be fully split up into linear factors over R, then there exists a basis of V so that the matrix of A is upper triangular.
For me to get to the point I don't understand that well I have to give part of the proof:
So suppose \lambda_{1} is an eigenvalue of A and v_{1} an eigenvector with this eigenvalue. Then W_{1} := <v_{1}> is an A-invariant subspace of V.
Expand v_{1} to a basis v_{1}, v_{2},..,v_{n} of V. Then W_{2}:=<v_{2},..,v_{n}> and V = W_{1} \oplus W_{2}
The matrix of A will be \left( \begin{array}{cc} \lambda_{1}&*\\ 0&R \end{array}\right)
Now suppose A_{R} is the linear transformation of W_{2} with matrix R with regard to basis v_{2},..,v_{n}
The proof now says that for all w \epsilon W_{2}: A(w) - A_{R}(w) \epsilon W_{1}.
I don't understand why this is the case, and I have no idea why this is important for the completion of the proof because it's not explicitly mentioned further on.
Can someone help me out here? Thanks!