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 \(T\).
Solution
Let \(\alpha\) be a cyclic vector for \(T\), and put \(n=\dim V\). Then \[ \alpha,T\alpha,\ldots,T^{n-1}\alpha \] is a basis of \(V\) [@HK71].
1. Express \(U\alpha\) as a polynomial in \(T\) applied to \(\alpha\).
Because the cyclic vectors above form a basis, there are scalars \(a_0,\ldots,a_{n-1}\) such that \[ U\alpha=a_0\alpha+a_1T\alpha+\cdots+a_{n-1}T^{n-1}\alpha. \] Define \[ p(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}. \] Then \[ U\alpha=p(T)\alpha. \]
2. Commutation propagates this equality to every cyclic-basis vector.
Since \(UT=TU\), induction gives \[ UT^k=T^kU \] for every integer \(k\ge0\). Hence, for \(0\le k<n\), \[ \begin{aligned} U(T^k\alpha) &=T^kU\alpha\\ &=T^kp(T)\alpha\\ &=p(T)T^k\alpha, \end{aligned} \] because \(p(T)\) commutes with \(T\). Thus \(U\) and \(p(T)\) agree on each vector of the cyclic basis.
3. Conclude equality of the operators.
Two linear operators that agree on a basis agree on all of \(V\). Therefore \[ \boxed{U=p(T)}. \] So every operator commuting with a cyclic operator is a polynomial in that operator.