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Let $T$ be a linear operator on the finite dimensional vector space $V$. Suppose $T$ has a cyclic vector. Prove that if $U$ is any linear operator which commutes with $T$, then $U$ is a polynomial in $T$.

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marked as duplicate by Marc van Leeuwen linear-algebra May 9 at 12:22

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  • $\begingroup$ This is Exercise 8 of Section 7.2 in Linear Algebra, second edition, by Hoffman and Kunze. $\endgroup$ – Maurice P Jan 6 at 4:02
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    $\begingroup$ I've marked this as a duplicate to a newer but more complete question (after adding what I think is a concise but complete answer to the part involved here); the reason is mainly to link together obviously related questions. $\endgroup$ – Marc van Leeuwen May 9 at 12:24
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Let $v$ be a cyclic vector for $T$. Show that there exists a polynomial of degree at most $\dim(V)-1$ such that $Uv=P(T)v$. (Expand $Uv$ in the cyclic basis.) Then use $UT^k=T^kU$ to show that $U$ and $P(T)$ coincide on the cyclic basis of $V$.

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  • $\begingroup$ Nice solutions.But how to use commutativity to both U and P(T) are identical? $\endgroup$ – MathLover Nov 7 '18 at 6:15
  • $\begingroup$ @SRJ For all $k \in \{0, \ldots, \dim(V)-1\}$: $UT^k v = T^k Uv = T^k P(T) v = P(T) T^k v$. So $U$ and $P(T)$ coincide on the basis $(v, Tv, T^2v, \ldots, T^{\dim(V)-1}v)$ of $V$. $\endgroup$ – WimC Nov 7 '18 at 12:34
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If $v$ is a cyclic vector, then $U$ is determined just by its image $U(v)$ of$~v$, since commutation implies the relation $U(T^k(v))=T^k(U(v))$ that fixes it on other elements of the spanning set $\{\,T^k(v)\mid k\in\Bbb N\,\}$ of $V$. With $n=\dim(V)$ we can write $U(v)=\sum_{0\leq k<n}c_kT^k(v)$ and then $P=\sum_{0\leq k<n}c_kX^k$, and therefore have $P[T](v)=U(v)$; but then $P[T]$ is an operator commuting with $T$ and having the same image of$~v$ as $U$ does, so $U=P[T]$ by our opening sentence.

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