some detail
the proof is also quite easy. the point is that every space is spanned by T cyclic sequences, and any T cyclic sequence is independent, but it may not be true that a union of T cyclic sequences is independent.
so the main point of the proof is to show that a union of T cyclic sequences, whose spans may overlap, can be rearranged so that they do not over lap.
e.g. if (x,Tx,...T^a-1 x; y,Ty,...,T^b-1 y) spans V, and a > b-1, we want to replace y by a vector of form y' = y - P(T)x, so that the union of T cyclic sequences
(x,Tx,...T^a-1 x; y', Ty',...,T^c-1 y') is a basis for V, where c-1 < b.
so if the dimension of V is a+c, where c-1< b, we need a polynomial P such that T^c(y-P(T)x) = 0. But since T^b y = 0, it follows (by the division algorithm) there is some c: c-1 < b, with T^c y in the span X of (x,Tx,...T^a-1 x), and such that for any other poly Q, then Q(T)y is in X iff T^c divides Q.
now since T^c y belongs to X, there is some poly Q such that Q(T)x = T^c y, and thus
0 = T^b y = T^(b-c) T^c y = T^ (b-c) Q(T) x. hence since T^a x = 0 and this is the minimum power that has this property, it must be true again by the division algorithm that T^ a divides T^(b-c) Q(T), i.e,. since b > c-1, that T^(a-b+c) divides Q(T).
but a-b >-1, so T^c divides Q(T). thus we have T^c y = Q(T) x = T^c P(T) x.
hence if we take y ' = y - P(T)x, we get T^c y' = 0.
then the sequences (x,Tx,...T^a-1 x; y' Ty',...,T^c-1 y') still span V and are independent.
'this is the key inductive step. next if V is spanned by several such sequences:
(x,Tx,...T^a-1 x; y, Ty,...,T^b y; ...; z,Tz,...T^c z), then
by induction V/X is a direct sum of T cyclic sequences, spanned by y', ...,z', and their nonzero T images.
then considering each pairwise span X + span(y', Ty',...);...; X+span(z',Tz',...) separately, we get a decomposition of V into a direct sum of X and subspaces isomorphic to the ones decomposing V/X.