Why a polynomial that annulls a linear operator is a multiple of a minimal polynomial? I was seeing the proof that a polynomial that annul in a linear operator is a multiple of a minimal polynomial, but I don't understand very well this demonstration:
Let $T\in L(V,V)$ a linear operator. If $p(x) \in \mathbb{P}(\mathbb{K})$ is a polynomial such that $p(T)(v)=0$, for all $v\in V$, so $p(x)$ is a multiple of $m_T(x)$. Indeed, do the division of $p(x)$ by $m_T(x)$, in other words, write $p(x)=m_T(x)*q(x)+r(x)$, where $r(x)=0$ or $\deg(r(x)) < \deg(m_T(x))$. Suppose that $r(x)\neq0$, so $r(x)=\sum_{i=0}^{s}b_i x^i$ with $b_s\neq0$ and $s=\deg(r(x)) < \deg(m_T(x))$. For $v \in V$, we have
$0 = p(T)(v) = (m_T * q)(T)(v) + r(T)(v) = (m_T(T) \circ q(T))(v) + r(T)(v)$
because $m_T(T)$ and $q(T)$ are polynomials in $T$, they commute and then
$(m_T(T) \circ q(T))(v) + r(T)(v)=(q(T) \circ m_T(T))(v) + r(T)(v)=q(T)  (m_T(T))(v) + r(T)(v)=r(T)(v)$, therefore
$\sum_{i=0}^{s}b_i T^i(v)=0$, for all $v \in V$ and then $T^s(v)=- \sum_{i=0}^{s} \frac{b_i}{b_s} T^i(v)$, for all $v \in V$, so {$T^0, T, ..., T^s$} is linearly dependent, a contradiction with the definition of $m_T(x)$, so $r(x)=0$ and the result is proven
My doubt is how the fact that $m_T(T)$ and $q(T)$ are polynomials in T ensures that they commute and how the author of the book claim this $(m_T(T) \circ q(T))(v) + r(T)(v)=(q(T) \circ m_T(T))(v) + r(T)(v)=q(T)  (m_T(T))(v) + r(T)(v)$?
 A: This proof becomes more transparent if you alleviate the notation. To begin with, I will drop the silly "$(x)$" that systematically follows each name of a polynomial; if something (usually $T$) gets substituted for $x$ I'll write that in square brackets (to avoid possible confusion with multiplication by a parenthesised expression). All that is used about the set $S=\{\,p\in K[x]\mid p[T]=0\,\}$ of polynomials annulled by the substitution $x:=T$, is that $S$ is a subgroup of $K[x]$ for addition, and that it is closed under multiplication by arbitrary polynomials: if $p\in S$ and $q\in K[x]$ then $pq\in S$. In ring-theoretic language, $S$ is an ideal of the commutative ring $K[x]$ of polynomials in $x$. Both mentioned properties follow easily from the definition and known properties of substitution, for instance for the second property one has $(pq)[T]=p[T]\circ q[T]=0\circ q[T]=0$ so $pq\in S$.
Now the argument is as follows. By definition $m_T$ is a nonzero (in fact monic) element of $S$ of the lowest possible degree. If $p\in S$ is arbitrary, then do Euclidean division of $p$ by $m_T$ to get $p=m_Tq+r$ with $\deg(r)<\deg(m_T)$. Now since $m_T\in S$ one has $m_Tq\in S$ (second property) and since $r=p-m_Tq$ one has $r\in S$ (first property). By minimality of $m_T$, this implies $r=0$, completing the proof that $m_T$ divides$~p$.

This is a very classical proof that every nonzero ideal in $K[x]$ has a nonzero element that divides all elements of the ideal (any nonzero element of minimal possible degree will do, and is called generator for the ideal), in other words that $K[x]$ is a principal ideal domain (PID). It is also used to prove that that the least common multiple of two polynomials $p,q$ divides any common multiple of them, and also to prove that their $\gcd$ can be written as a $sp+tq$ for certain (Bézout coefficients) $s,t\in K[x]$ (the latter proof actually amounts to proving that a generator of the ideal of combinations $sp+tq$ is a common divisor of $p$ and $q$).

Finally I'll answer your actual questions about the proof in the book. The commutation of polynomials in $T$ is a consequence of substitution of $T$ for $x$ being compatible with multiplication/composition, namely $(pq)[T]=p[T]\circ q[T]$ for any $p,q\in K[x]$, since this same rule can be applied to $qp$, which his equal to$~pq$. Actually you wouldn't need commutation in the cited proof, for two independent reasons: (1) if one had written $p=qm_T+r$ rather than $p=m_Tq+r$, then the operator $m_T[T]$ would have appeared on the right from the beginning, and (2) even with $m_T[T]$ on the left, one can immediately conclude that $m_T[T](q[T](v))=0$ since the outermost operator applied is null. For you final question, it is not clear what is difficult about the equations $(m_T[T] \circ q[T])(v)=(q(T) \circ m_T[T])(v)=q(T]  (m_T[T])(v)$ (where I've dropped the passive term $r[T](v)$), as the first just applies commutation, and the second the definition of function composition. Maybe you meant to ask why the term $q[T](m_T[T](v))$ is zero, that is because $m_T[T](v)=0$ (since $m_T[T]=0$) and $q[T]$ is a linear operator (so it sends $0$ to $0$).
A: The first fact (that $P(T)$ and $Q(T)$) commute if $P$ and $Q$ are polynomials) come from the application :
$$
\begin{array}{c}
\Psi_T : & \mathbb{C}[X] & \rightarrow & L(V) \\
 &\sum_{i=0}^n a_i X^i & \mapsto & \sum_{i=0}^n a_i T^i
 \end{array}
$$
Is a morphism of algebra.
In fact :
$$
P(T)\circ Q(T)= \Psi_T(P)\circ\Psi_T(Q)=\Psi_T(PQ)=\Psi_T(QP)=\Psi_T(Q)\circ\Psi_T(P)=Q(T)\circ P(T)
$$
For the second question, the first equality come from the commutativity shown now, and the second is a simple use of definition of $f\circ g(x)=f(g(x))$. 
