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Triangulization theorem

  1. Jul 28, 2005 #1
    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!
  2. jcsd
  3. Jul 28, 2005 #2

    matt grime

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    just work it out: what is A-A_r with respect to that basis?
  4. Aug 1, 2005 #3
    Okay..I have to substract 2 transformations then. How do I do that? I know that per definition (f+g)(x) = f(x) + g(x), but what are f(x) and g(x) in this case?

    I can see the coordinates of both bases, but they have a different number of coordinate numbers, right? So I can't substract them..

    Don't see it. :frown:
  5. Aug 1, 2005 #4
    Oh wait, if I use [tex]A_{r}[/tex] on [tex]w[/tex] [tex]\epsilon[/tex] [tex]W_{2}[/tex], I have to convert that w to its coordinates with respect to the basis of [tex]W_{2}[/tex].

    Then these n-1 coordinate numbers match the last n-1 coordinate numbers of A(w) with respect to the basis of V, is that right?

    So if I substract these coordinates, I get a coordinate vector with only the first coordinate number possibly not zero, so that is an element of [tex]W_{1}[/tex].

    Is that the way to see it?
  6. Aug 3, 2005 #5
    Okay..assuming I'm right, can anyone tell me why this is important for the proof? Because I can't see it.

    Proof continued (proof by induction on the dimension):
    By calculating [tex]|tI_{p} - A|[/tex] using cofactor expansion for the first column we get [tex]f_{A}(t) = (t - \lambda_{1}) f_{A_{R}}(t)[/tex].

    Thus [tex]f_{A_{R}}(t)[/tex] also fully splits into linear factors over R (real numbers).

    The induction hypothesis, applied on the (p-1)-dimensional vectorspace [tex]W_{2}[/tex] and the linear transormation [tex]A_{R}[/tex], now gives a basis [tex]v_{2}',..,v_{n}'[/tex] of [tex]W_{2}[/tex] so that the matrix R' of [tex]A_{R}[/tex] with respect to this new basis is upper triangular.

    The vectors [tex]v_{1},..,v_{n}'[/tex] now also form a basis of V and the matrix of A with respect to this basis is of the form

    [tex]\left( \begin{array}{cc} \lambda_{1}&*'\\ 0&R' \end{array}\right)[/tex]

    which is upper triangular. QED

    Thanks in advance.
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