Finding the generator for a cyclic module

Asked Jan 04 '21 at 21:16

Active Jan 04 '21 at 21:16

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Let $V$ be a finite dimensional vector space over $F$, and $T:V \to V$ a linear transformation. We all know that $V$ can be considered an $F[x]$-module. Moreoever, we know that if the minimal polynomial is equal to the characteristic polynomial, then $V$ is actually a cyclic $F[x]$-module. This follows immediately from invariant decomposition.

My question is, do we have an explicit way to find the generator $v \in V$?

From the above blurb, we know that $V \cong F[x]/(m_T(x))$ where $m_T(x)$ is the minimal polynomial of $T$. So, in order to find $v$, we just need to find the element in $V$ that maps to $\bar{1} \in F[x]/(m_T(x))$ under that isomorphism. But what is our isomorphism?

asked Jan 04 '21 at 21:16

Good Morning Captain

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For what it's worth, a "randomly chosen" vector $v$ will be a cyclic vector. – Ben Grossmann Jan 04 '21 at 22:20
@BenGrossmann So any vector will generate $V$ as an $F[x]$-module? – Good Morning Captain Jan 04 '21 at 22:22
No, but "almost all vectors" will generate $V$ as an $F[x]$ module – Ben Grossmann Jan 04 '21 at 22:23
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In this post, I give a complete characterization of the cyclic vectors (i.e. minimal generators of the module) in terms of the Jordan normal form. – Ben Grossmann Jan 04 '21 at 22:25
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Another way of putting it is that the set of vectors that fail to generate the module form a union of finitely many proper subspaces of $V$. If $V$ is identified with $\Bbb Q^n, \Bbb R^n,$ or $\Bbb C^n$, then this set of "bad" vectors has Lebesgue measure zero. – Ben Grossmann Jan 04 '21 at 22:29
@BenGrossmann I will read that post. Thanks! – Good Morning Captain Jan 04 '21 at 22:34

Finding the generator for a cyclic module

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