OpenStudy (anonymous):

A little question about Jordan Canonical Form

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago