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Let $\alpha$ and $\beta$ be linear operators on a vector space V. Suppose that $\alpha$ satisfies a non-zero polynomial. Prove that when V is $\alpha$- cyclic, we have $\alpha \circ \beta = \beta\circ\alpha,$ then $\beta = p(\alpha)$ for some polynomial p.

My approach was to let the $p'$ be the minimal polynomial that satisfies alpha. Then let $p'(\alpha\circ\beta) = p'(\beta\circ\alpha)$ but I can't seem to proceed from here. I think the next step I want to show is that there is a polynomial that satisfies $\beta$, but after that I am just stuck

TUFFCLUBBOY
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  • What does $\alpha$-cyclic mean in this context? Also what do you mean by "minimal" polynomial; minimal degree? – Spencer May 05 '19 at 00:19
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    @Spencer In linear algebra the minimal polynomial of a linear transformation $f$ is the monic polynomial $p$ of smallest degree such that $p(f)$ is the zero linear transformation. [The tag above mentions the definition] A vector space is $f$-cyclic when there is a vector $v\in V$ such that $v,f(v),f^2(v),...$ span $V$. – logarithm May 05 '19 at 00:33
  • see http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 – Will Jagy May 05 '19 at 01:50

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