Hoffman Kunze, Lemma concerning triangulable operators I am currently studying Linear Algebra by myself and recently I have stumbled across an important Lemma in Hoffman Kunze's book:

Lemma. 
  Let $V$ be a finite-dimensional vector space over the field $F$. Let $T$ be linear operator on $V$ such that the minimal polynomial for $T$ is a product of linear factors
$$p = (x-c_{1})^{r_{1}}...(x-c_{k})^{r_{k}}, c_{i} \in F.$$
Let $W$ be a proper subspace of $V$ which is invariant under $T$. there  exists a vector $\alpha$ in $V$ such that
a) $\alpha\notin W$;
b) $(T-cI)\alpha\in W$, for some characteristic value $c$ of the operator $T$.

I must admit that I'm a little confused at this point. If I understand correctly, by using this Lemma we can produce a sequence $\{0\} = W_{0} \subset W_{1} \subset \ldots \subset W_{n} = V$ of invariant subspaces for $T$ s.t. $\dim{W_{i}}=i$, which would give me a basis in which the matrix of the operator $T$ is triangular. But this is not clear to me. Can someone explain to me the exact meaning of the condition $(b)$? Why this $\alpha$, together with the old basis, spans an invariant subspace? How does it all relates to the invariant hyperplanes - are those exactly what $W_{i}$'s are?
The book says that $(a)$ and $(b)$ state that the $T$-conductor of $\alpha$ into $W$ (i.e. the unique monic generator of the ideal $S(\alpha ; W)$) is a linear polynomial - that is also a little confusing to me and I would appreciate some explanation of this statement. Thanks in advance for any help.
 A: Let's tackle the second part first. By definition, the $T$-conductor of $\alpha$ into $W$ is the smallest degree monic polynomial $q$ where $q(T)\alpha\in W$, i.e. $q$ is the monic generator of $S(\alpha;W)$. By $(b)$, $q$ divides $x-c$ since $S(\alpha;W)$ is a polynomial ideal. By $(a)$, $q$ is not a nonzero constant. $($Suppose that $q(x) = k$ for some $k\in F$. Note that $q$ is monic and therefore $k=1$. Then $\alpha = I\alpha = q(T)\alpha\in W$, which contradicts $(a).)$ Therefore $q$ must be $x-c$.
Now let's examine the conclusions of the lemma. Rewriting $(b)$, we have that
$T\alpha = c\alpha + \beta$ for some $\beta\in W$. $W$ is $T$-invariant, so this guarantees that $W^\prime = \operatorname{span}\{\alpha\} \oplus W$ is also $T$-invariant (where $\oplus$ denotes the direct sum of subspaces). This is what allows the subsequent construction of a basis that triangulates $T$.
We start with $W_0 = \{0\}$. By the lemma, there is an $\alpha_1 \notin W_0$ where $T\alpha_1\in W_1 = \operatorname{span}\{\alpha_1\} \oplus W_0  =\operatorname{span}\{\alpha_1\}$ and $W_1$ is $T$-invariant with dimension $1$.
Now we proceed inductively. At each stage $i\leq n$ we use the lemma to choose a new $\alpha_i$ so that we have a subspace $W_i = \operatorname{span}\{\alpha_i\}\oplus W_{i-1} =\operatorname{span}\{\alpha_1, \dots, \alpha_i\}$ where


*

*$\alpha_i\notin W_{i-1}$, but $T\alpha_i\in W_{i}$, 

*$W_i$ is $T$-invariant and has dimension $i$.
At the end we have a basis $\cal{B}=\{\alpha_1, \dots, \alpha_n\}$ with $T\alpha_i\in\operatorname{span}\{\alpha_1, \dots, \alpha_i\}$. This is precisely the condition on a basis that guarantees $[T]_\cal{B}$ is upper triangular.
