Why is A(w) - A_R(w) in W_1 in the Triangulization Theorem?

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In summary, the linear transformation A has a characteristic polynomial f_{A} which can be fully split up into linear factors over the real number R. This implies that there exists a basis of V so that the matrix of A is upper triangular. This is important for the proof of the theorem because it allows for the use of induction on the dimension.
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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!
 
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  • #2
just work it out: what is A-A_r with respect to that basis?
 
  • #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:
 
  • #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?
 
  • #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.
 

1. What is the Triangulization Theorem?

The Triangulization Theorem is a mathematical theorem that states that any polygon can be divided into triangles by drawing diagonals between its vertices.

2. Who discovered the Triangulization Theorem?

The Triangulization Theorem was discovered by Leonhard Euler, a Swiss mathematician, in the 18th century.

3. What is the significance of the Triangulization Theorem?

The Triangulization Theorem has many practical applications, such as in computer graphics, where it is used to break down complex shapes into simpler ones for easier rendering.

4. Can the Triangulization Theorem be applied to any polygon?

Yes, the Triangulization Theorem can be applied to any polygon, regardless of its size or shape. This is because any polygon can be broken down into a series of triangles.

5. Are there any limitations to the Triangulization Theorem?

The Triangulization Theorem does have some limitations, such as not being applicable to polygons with holes or self-intersecting polygons. In these cases, the polygon may need to be divided into multiple sections to be triangulated.

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