(1)
By taking a maximal linearly independent subset of $\mathcal{F}$ means that every other function in $\mathcal{F}$ can be expressed as a linear combination of those functions in the maximal set, and of course condition (b) is invariant under linear combinations (if $f_1$, $f_2$ satisfy (b) then their linear combination satisfies (b)). Also -for some reason that I cannot see at the moment, and I hope it's not the main question- this maximal set can be considered finite.
(2)
I suppose you meant "$V_1$ larger than $W$". Now if $x\in W$, then $T_1(x)\in W$ as $W$ is invariant under $\mathcal{F}$. Hence $T_1(x)-c_1I(x)=T_1(x)-c_1x$ is a linear combination of elements of $W$, and thus is in $W$. So every $x\in W$ is in $V_1$.
(3)
I would suggest you to take a closer look. It doesn't say that the statements are equivalent. He proves that If $T$ commutes with $T_1$ (and that's the case for every operator in $\mathcal{F}$, by definition) then $T(\beta)$ is in $V_1$ for any $\beta \in V_1$. In other words $V_1$ is invariant under $\mathcal{F}$ (keep that in mind for your fourth question)
(4)
$V_2$ was defined for $T_2$ as $V_1$ was defined for $T_1$ with the additional property that ot is a subspace of $V_2$. With the same reasoning as in (3) (and also using (3)), we can show that for $T(\beta)\in V_2$ for any $\beta \in V_2$, for all $T\in \mathcal{F}$. In other words (again) $V_2$ is invariant under $\mathcal{F}.$
Edit: Question (1)
It's finite because $\dim V=n<+\infty$, for some $n\in \mathbb{N}$. Every linear operator $T$ can be represented by one $n\times n$ matrix. Every euch matrix is a linear combination of (the linear independent) matrices of the form:
- A_{ij} = $a_{ij}=1$ and every other $a_{i' j'}=0$, $i'j'\neq ij$, $ \ $ where $i,j,i',j'\in \{1,2,\ldots , n\}$.
Of course the set of all such matrices is finite.
Well, (2) implies (1) as for $T$ commutes with itself, so about (2): For $i = 1, …, k$, as $W_i = \ker p_i^{r_i}(T)$,
$$(p_i^{r_i}(T)∘U)(W_i) = (U∘p_i^{r_i}(T))(W_i) = U(p_i^{r_i}(W_i)) = U(0) = 0,$$
so $U(W_i) ⊆ \ker p_i^{r_i}(T) = W_i$.
Best Answer
(1) Of course, $c_j$ is a root of $q-q(c_j)$. Hence there is some polynomial $h$ such that $q-q(c_j) = (x-c_j)\cdot h$.
(2) Since $p$ is the minimal polynomial of $T$, one has $p(T) = 0$ by definition.
Moreover, $W$ was defined to be the span of all characteristic vectors of $T$. Here, one has $0 = (T-c_j I)q(T)\alpha$. In other words, $q(T)\alpha \in \mathrm{ker}(T-c_jI)$, which means that $q(T)\alpha$ is a characteristic vector of $T$ to the eigenvalue $c_j$ (or $q(T)\alpha = 0$).