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

[TomMe]

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!

 

[matt grime]

just work it out: what is A-A_r with respect to that basis?

 

[TomMe]

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.

 

[TomMe]

Oh wait, if I use $$A_{r}$$ on $$w$$ $$\epsilon$$ $$W_{2}$$, I have to convert that w to its coordinates with respect to the basis of $$W_{2}$$.

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 $$W_{1}$$.

Is that the way to see it?

 

[TomMe]

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):
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By calculating $$|tI_{p} - A|$$ using cofactor expansion for the first column we get $$f_{A}(t) = (t - \lambda_{1}) f_{A_{R}}(t)$$.

Thus $$f_{A_{R}}(t)$$ also fully splits into linear factors over R (real numbers).

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

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

$$\left( \begin{array}{cc} \lambda_{1}&*'\\ 0&R' \end{array}\right)$$

which is upper triangular. QED
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Thanks in advance.