# T-cyclic subspaces: If $m_v$ is irreducible, show that $C_T(v) = C_T(w)$ for all $w\in C_T(v)$

We consider a linear transformation $T\in$ End$(V)$ and $V$ vector space (not necessarly finite).

Notation:

$\mathcal{C} = v, T(v), T^2(v), ...$

$C_T(v)$ = Span$(\mathcal{C})$

$m_v$ is the minimal polynomial of $v$ related to $T$.

The question itselft is:

Let $p\in F[x]$ be a irreducible polynomial and suppose that $m_v = p$ for some $v\in V$. Prove that $C_T(w)=C_T(v)$ for all $w\in C_T(v)$.

The only idea that i had was to take $w\in C_T(v)$ and try to relate $f\in F[x]$ s.t $f(T)= w$ with $m_v$. But i dunno exactly how.

Any hint?

By definition:$w=Q(T)(v)$, with $Q$ some polynomial. We may assume without loss of generality, that $deg(Q)<deg(p)$, simply by taking a long division of $Q$ by $p$. Since $p$ is irreducible, either $Q=0$ or $p,Q$ are coprime.
$Q$ cannot be zero, since by assumption $w\neq 0$. Hence $p$ and $Q$ are coprime, therefore there exist polynomials $U,V$ such that $pU+QV=1$, therefore $pU(T)(v)+QV(T)(v)=v$. $pU(T)(v)=0$ by definition of $p$ and $Q(T)(v)=w$, hence $V(T)(w)=v$.
From this it followis trivially that $C_T(w)=C_T(v)$.
• Thank you. In order to justify the wlog part, do you agree with this argument? > If $deg(Q)>deg(p)$, we take a long division of $Q$ by $p$. Then, $Q$ can be written "in base $p$", that is, $Q = q_k.p^k+...+ r_2.p+r_1$, where $deg(q_k),deg(r_i)<deg(p)$. By the argument you gave, for $q_k$ and for every $r_i$ there corresponds a $v_k$ and $v_i$ such that $C_T(v_i), C_T(v_k) = C_T(v)$. By the T-invariancy, therefore, $C_T(w)$ = $C_T(v)$ Aug 19 '17 at 13:01