# $T$-annihilators divide minimal polynomial of a linear operator $T$

In Hoffman and kunze linear algebra Sec.7.1 , p-228.

If $T$ be a linear operator on a finite dimensional vector space$V$ over a field $F$.

I can understand how minimal polynomial of a linear operator $T$ belongs to the ideal $T$-annihilator of $\alpha$ , for every $\alpha \in V,$ as $M (\alpha ;T)$.

But I can't understand how every element of the ideal $T$-annihilator of $\alpha$ divide the minimal polynomial of $T$.

the confusion started from p-202 , sec-6.4 ,

where in the case of $T$-conductor of $\alpha$ into $W$ , $S (\alpha; W)$ . In Hoffman and kunze it's written that " every $T$-conductor divides the minimal polynomial for $T$".

Please help.

## 1 Answer

Every element of the $T$-annihilator of $\alpha$, $M(\alpha; T)$, does not divide the minimal polynomial $m_T$ of $T.$ For example, $m_T^2\in M(\alpha; T)$. It's the generator $p_\alpha$ of $M(\alpha; T)$ that divides $m_T$, which is clear since $m_T\in M(\alpha; T)$ and $p_\alpha$ divides every element of $M(\alpha; T)$. I think you are getting confused because Hoffman and Kunze call both $M(\alpha; T)$ and $p_\alpha$ the $T$-annihilator of $\alpha$; you'll have to pay attention to context to determine whether they mean the ideal or the generator in any particular argument.

• I noticed that and think the same as your comment but the confusion started from p-202 , sec-6.4 , where in the case of $T$-conductor of $\alpha$ into $W$ , $S (\alpha; W)$ . In Hoffman and kunze it's written that " every $T$-conductor divides the minimal polynomial for $T$". – Sandip Agarwal Apr 19 '18 at 3:46
• They're using the same convention here as well. The $T$-conductor of $\alpha$ into $W$ refers to both the ideal $S(\alpha;W)$ and its monic generator. In the context you're quoting, they mean the generator. – Christian Sykes Apr 19 '18 at 3:52
• Thanks, for clearing the confusion. – Sandip Agarwal Apr 19 '18 at 3:54