# Cyclic vector implies commuting linear operator is a polynomial [duplicate]

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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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 $V$.
• @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 $V$. – WimC Nov 7 '18 at 12:34
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 $$U(v)=\sum_{0\leq k and then $$P=\sum_{0\leq 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.