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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!

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  • $\begingroup$ I think that in (A) vector $w$ is what you need. $v$ will not satisfy (A) unless $r=1$. $\endgroup$
    – richrow
    Jun 20, 2021 at 8:18
  • $\begingroup$ @richrow : you are completely right. Edited. Thank you! $\endgroup$ Jun 20, 2021 at 8:24
  • $\begingroup$ 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). $\endgroup$
    – richrow
    Jun 20, 2021 at 8:25
  • $\begingroup$ @richrow: if it's not too rude to ask - will you please be able to help me out with proving part (B)? Thank you! $\endgroup$ Jun 20, 2021 at 10:54

1 Answer 1

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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 $q=min_{u,T}min_{w,T}$

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  • $\begingroup$ thank you for the answer, can you please explain what do you mean by "as $q$ annihilates? $\endgroup$ Jun 20, 2021 at 17:40
  • $\begingroup$ a polynomial $p$ is an annhilator of $T$ if $p(T)=0$, i.e. $p(T)$ is the zero operator. $\endgroup$
    – AGH
    Jun 21, 2021 at 5:15

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