# $\overline T:V/W \rightarrow V/W$ and $m_{\overline T}$ its minimal polynomial. Show that $m_T = m_Sm_{\overline T}$

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?

• $m_{\overline T},m_v$ don't have to be relatively prime – reuns Feb 24 '20 at 23:12

$$f(v)= m_S(T) v + W$$ 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 $$m_T= m_Sm_\overline{T}$$.
• 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. – Marc van Leeuwen Feb 29 '20 at 20:38