- #1

TomMe

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Suppose [tex]A: V -> V[/tex] a linear transformation, with dim(V) = n. If the characteristic polynomial [tex]f_{A}[/tex] 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 [tex]\lambda_{1}[/tex] is an eigenvalue of A and [tex]v_{1}[/tex] an eigenvector with this eigenvalue. Then [tex]W_{1} := <v_{1}>[/tex] is an A-invariant subspace of V.

Expand [tex]v_{1}[/tex] to a basis [tex]v_{1}, v_{2},..,v_{n}[/tex] of V. Then [tex]W_{2}:=<v_{2},..,v_{n}>[/tex] and [tex]V = W_{1} \oplus W_{2}[/tex]

The matrix of A will be [tex]\left( \begin{array}{cc} \lambda_{1}&*\\ 0&R \end{array}\right)[/tex]

Now suppose [tex]A_{R}[/tex] is the linear transformation of [tex]W_{2}[/tex] with matrix R with regard to basis [tex]v_{2},..,v_{n}[/tex]

The proof now says that for all [tex]w[/tex] [tex]\epsilon[/tex] [tex] W_{2}: A(w) - A_{R}(w)[/tex] [tex]\epsilon[/tex] [tex]W_{1}[/tex].

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!