I am currently a year 2 uni mathematician, understanding the Jordan Normal Form in Linear Algebra.
Here is a paper for a short proof of the Jordan Normal Form, which I am looking at in order to understand it:
http://www.ma.rhul.ac.uk/~uvah099/Maths/JNFfinal.pdf
It reduces the proof down to the following statement:
If $T: V \to V$ is a linear transformation of a finite-dimensional vector space such that $T^m = 0$ for some $m \geq 1$, then there is a basis of $V$ of the form
$u_1, T(u_1), \dots, T^{a_1 - 1}(u_1), \dots, u_k, T(u_k), \dots, T^{a_k - 1} (u_k)$
where $T^{a_i}(u_i) = 0$ for $1 \leq i \leq k$.
I am currently struggling to understand the proof, and how the adjoined elements $w_1, \dots, w_m$ are in fact of the form described in the theorem. That is, how can we show that $w_1, \dots, w_m$ are also something like $v,T(v), \dots, T^{m-1}(v)$?
Any help would be greatly appreciated.
Let us call $v, T(v), \ldots, T^{m-1}(v)$, where $T^m(v) = 0$, the cycle generated by $v$. The point is to get a basis that can be viewed as a union of disjoint cycles. In the proof you linked the vectors $w_1, \ldots, w_m$ are all taken from $\ker(T)$, so they have $T(w_i) = 0$ for all $i$. Another way of saying that is that the cycle generated by $w_i$ is just $w_i$ itself, so we're just adding a bunch of cycles of length $1$ here.