Question

A little question about Jordan Canonical Form

Answer 1

I'm finishing studying it, I understood all proofs in my book, but it only says:
http://imageshack.us/photo/my-images/41/jordanzm.png/
An attemp of translation:
Consider B the basis of V obtained resorting this basis as follows:
(...) (basis)
You can verify that \[\left| f \right|_B\] has the form of the statament of the theorem.

It's in the page 167 (It's in Spanish :()
http://mate.dm.uba.ar/~jeronimo/algebra_lineal/AlgebraLineal.pdf


I can understand it if I find a proof that a nilpotent block A in K nxn such as \[A^{k} = 0\] and \[A^{k-1} \neq 0\], if we choose a v that isn't the kernel of A^{k-1}, and we make a basis:
{v,Av,A^2v,...,A^(k-1)v} then
\[\left| f \right|_B\] =
http://upload.wikimedia.org/math/f/f/b/ffb7cdd732e1575c2f1ed9f37d222dac.png
(I understand the rest)

PD: In my book jordan form has the ones in the subdiagonal instead the supperdiagonal, but it's the same, I think that you only have to sort the vectors of the base in the opposite way to achieve the other definition of jordan canonical form.

Answer 2

PD: I think I have to study technical English as soon as possible xD

Answer 3

I was about to post this:
[b]1. The problem statement, all variables and given/known data[/b]
I was studying the existence of a jordan canonical form, and I understood the proofs but there was something unproved:
Translated (The original text is in Spanish)
Consider the basis B of V, obtained resorting this base as follows:
\[B = \left \{ v^{(k)}_1, f(v^{(k)}_1), ... , f^{k-1}(v^{(k)}_1),...,v^{(k)}_{r_{k}},f(v^{(k)}_{r_{k}}),...,f^{k-1}(v^{(k)}_{r_{k}}), ... ,v^{(j)}_{1}, f(v^{(j)}_{1}),...,f^{j-1}(v^{(j)}_{1}),...,v^{(j)}_{r_{j}},f(v^{(j)}_{r_{j}}),...,f^{(j-1)}(v^{(j)}_{r_{j}}),...,v^{(1)}_{1},...,v^{(1)}_{r_{1}} \right \}\]
I think I can understand this, assuming that:
Let's be f:V->V a nilpotent linear transformation such as:
m_f=X_f=X^k
(m_f is the minimal polynomial and X_k is the charasteristic polynomial)
Then
\[B = \left \{ v,f(v),f^{2}(v),...,f^{k-1}(v) \right \}\] it's a basis of V (I know it's, because k is the dimension of V because X_f=X_k and if it wasn't linear independent, the minimal polynomial would have a grade lesser than k).
So:

\[\left | f \right |_{B}=
\begin{pmatrix}
0 &0 &... &0 &0 \\
1 &0 &... &0 &0 \\
0 &1 &... &0 &0 \\
... &... &... &... &... \\
0 &0 &... &1 &0
\end{pmatrix}\]
This isn't proven either.

----

In a forum and I realized how to do this (and it was pretty obvious but sometimes I fail to see the obvious, so it may not be so obvious as I'm saying xD).
So I'll close this.

Answer 4

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Answer 5

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Answer 9

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Answer 15

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Answer 18

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Answer 19

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Answer 20

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Answer 21

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Answer 23

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B.