# Proving the minimal polynomial is the minimal polynomial of one of the vectors [closed]

In the following question, we need to prove the following statement:

Let $$V$$ be a vector space over $$\mathbb{F}$$ such that $$dim(V)=n<\infty$$, and let $$T:V\to V$$ be a linear map. Then, there exists a vector $$v\in V$$ such that $$\mu_T (x) = min_{v,T}(x)$$ where $$\mu_T (x)$$ is the minimal polynomial of $$T$$, and $$min_{v,T}$$ is the least degree monic polynomial such that $$P(T)v=0$$.

The question is divided into three parts:

(A) Prove the statement for the case $$\mu_T(x)=P^r(x)$$ where $$P(x)\in \mathbb{F}[X]$$ is an irreducible polynomial and $$r\geq 1$$.

(B) If $$V=U \bigoplus V$$ where $$U,V$$ are T-invariant, such that $$min_{u,T}, min_{w,T}$$ are coprime, then $$min_{v,T}=min_{u,T}\cdot min_{v,T}$$ where $$v=u+w$$.

(C) Prove the general statement.

I will be happy to receive help in all three parts. Here are my ideas:

(A) $$P^r(T)v=0$$ for all $$v\in V$$. There must exist $$w\in V$$ for which $$P^{r-1}(T)w\neq 0$$ (otherwise, the minimal polynomial would be of smaller degree). If we take $$w$$ we obtain the required vector (is this correct?).

Is (C) the result of the first two statements, together with the primary decomposition theorem? I am not sure.

Thank you!

• I think that in (A) vector $w$ is what you need. $v$ will not satisfy (A) unless $r=1$. Jun 20, 2021 at 8:18
• @richrow : you are completely right. Edited. Thank you! Jun 20, 2021 at 8:24
• As for (C) - yes, you are right, it follows from the previous parts and primary decomposition. If $\mu_T=P^r\cdot Q$, then you have the decomposition $V=\ker P^r\oplus\ker Q$, so now you can apply (B). Jun 20, 2021 at 8:25
• @richrow: if it's not too rude to ask - will you please be able to help me out with proving part (B)? Thank you! Jun 20, 2021 at 10:54

For (B): We write $$v=u+w$$. Let $$min_{v,T}(x)=q(x)$$. Now, $$q(T)v=0=q(T)u+q(T)w$$ As $$U,V$$ are invariant under $$T$$, $$q(T)u\in U$$ and $$q(T)w\in V$$. Therefore, $$q(T)u=0=q(T)w$$. As $$q$$ annihilates, $$min_{u,T}$$ divides $$q$$, and $$min_{w,T}$$ divides $$q$$, therefore, as they are coprime, $$min_{u,T}min_{w,T}$$ divides $$q$$, and as $$min_{u,T}(T)min_{w,T}(T)v=0$$, we can conclude that $$q=min_{u,T}min_{w,T}$$
• thank you for the answer, can you please explain what do you mean by "as $q$ annihilates? Jun 20, 2021 at 17:40
• a polynomial $p$ is an annhilator of $T$ if $p(T)=0$, i.e. $p(T)$ is the zero operator.