Let \(T\) be a linear operator on the finite-dimensional space \(V\). Prove that \(T\) has a cyclic vector if and only if the following is true: Every linear operator \(U\) which commutes with \(T\) is a polynomial in \(T\).
Solution
1. A cyclic vector forces every commuting operator to be a polynomial in \(T\).
If \(T\) has a cyclic vector, Exercise 7.1.8 proves directly that every linear operator commuting with \(T\) is a polynomial in \(T\) [@HK71].
2. If \(T\) has no cyclic vector, construct a commuting operator that is not a polynomial in \(T\).
Apply the cyclic decomposition theorem with \(W_0=0\). Then \[ V=Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T), \] where the nonzero vectors \(\alpha_i\) have \(T\)-annihilators \[ p_r\mid\cdots\mid p_2\mid p_1. \] If \(r=1\), then \(\alpha_1\) is cyclic for \(T\). Thus under the present assumption \(r\ge2\).
Define \(U\) to be zero on \(Z(\alpha_1;T)\) and the identity on \[ Z(\alpha_2;T)\oplus\cdots\oplus Z(\alpha_r;T). \] Each summand is \(T\)-invariant, so \(UT=TU\).
Suppose \(U=q(T)\) for some polynomial \(q\). Since \[ q(T)\alpha_1=U\alpha_1=0, \] the annihilator property gives \(p_1\mid q\). Hence every \(p_i\) divides \(q\). Therefore \(q(T)\) annihilates every cyclic summand, so \(U=q(T)=0\) on all of \(V\). But \(U\) is the identity on the nonzero space \(Z(\alpha_2;T)\), a contradiction. Thus this commuting \(U\) is not a polynomial in \(T\).
3. Conclude the equivalence.
Step 1 proves the forward implication, while step 2 proves the contrapositive of the reverse implication. Therefore \[ \boxed{T\text{ has a cyclic vector}\iff \operatorname{Comm}(T)=F[T].} \]