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Let $V$ be a complex finite dimensional vector space and $T\in \mbox{End}(V)$. Given $v\in V$, the family of vectors $(T^{i}(v))_{i\geq 0}$ is linearly dependent, and therefore we may consider the smallest integer $m\geq 1$ such that this family become linearly dependent. This does define a monic polynomial $m_v$, the smallest degree monomial such that $m_v(T)(v)=0$. Define $W=\{w\in V: m_v(T)(w)=0\}$. Since $W$ is $T$-invariant, we may consider two new linear operators: $S:W\rightarrow W$ and the induced operator $\overline T:V/W\rightarrow V/W.$ Writing, respectively, $m_T, m_S$ and $m_{\overline T}$ for the minimal polynomial of $T$, of $S$ and of $\overline T$, prove that $m_T=m_Sm_{\overline T}$.

My attempt:

Since for all $\overline u\in V/W$ we have $\overline 0 = m_{\overline T}(\overline T)(\overline u)$ iff $m_{\overline T}(T)(u) \in W$, then for all $u\in V$ we have $m_v(T)m_{\overline T}(T)(u) = 0$ and therefore $m_T|m_vm_{\overline T}.$ Now we notice that $m_v|m_T$, since $m_v$ is the smallest degree polynomial responsable for killing $v$. Also, we notice for all $\overline u\in V/W$ that $m_T(\overline T)(\overline u) = \overline{m_T(T)(u)} = \overline 0 \ $ and therefore $m_{\overline T}|m_T$. So it is the case that $lcm(m_v,m_{\overline T})|m_T$. If I could prove that $m_{\overline T},m_v$ are relatively prime, I would be done. But i'm struggling here and can't proceed. Any insight?

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    $\begingroup$ $m_{\overline T},m_v$ don't have to be relatively prime $\endgroup$ – reuns Feb 24 '20 at 23:12
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$f(v)= m_S(T) v + W$ is a linear map $V\to V/W$ with kernel $W$, thus the rank formula says it is surjective $V\to V/W$, and hence $p(T) m_S(T)=0\in Hom(V,V/W)$ iff $p(T)=0\in End(V/W)$ iff $m_\overline{T}\ |\ p$.

And since $m_S$ divides $m_T$ then $m_T= m_Sm_\overline{T}$.

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  • $\begingroup$ I think I can understand what you mean, but as a teacher I would be picky: polynomials in $T$ like $p(T)m_S(T)$ or $p(T)$ lie in $\operatorname{End}(V)$, just like $T$ does; you cannot consider them in $\operatorname{Hom}(V,V/W)$ or $\operatorname{End}(V/W)$ just because you want to them to. Maybe you want to talk about induced morphisms, maybe you meant something else than what you wrote (like $p(\bar T)\circ f$ respectively $p(\bar T)$), but as it stands it is not quite right. $\endgroup$ – Marc van Leeuwen Feb 29 '20 at 20:38

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