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    Equivalent conditions for an operator to have a cyclic vector

    PR-4GQIZ

    A linear operator is cyclic exactly when its minimal and characteristic polynomials agree, equivalently its rational form has one block.Suggest a slogan

    Let \(k\) be a field, let \(V\) be a nonzero finite-dimensional \(k\)-vector space, and let \(T\colon V \to V\) be a \(k\)-linear map. Make \(V\) a \(k[x]\)-module by \(f(x) \cdot v \coloneqq f(T)v\), and let \(\min_T(x)\) be the minimal polynomial of \(T\). The following are equivalent:

    • \(V \cong k[x]/\gens{\min_T}\) as \(k[x]\)-modules.

    • \(V\) admits a cyclic vector: there is \(v \in V\) such that \(v, Tv, T^2v, \ldots\) span \(V\). For such \(v\), \(\min_T\) is the monic polynomial \(p\) of least degree with \(p(T)v = 0\).

    • \(V\) is a cyclic \(k[x]\)-module. Its annihilator is then the ideal \(\gens{\min_T}\).

    • The matrix of \(T\) in some basis of \(V\) is the companion matrix \(C_{\min_T}\).

    • \(\min_T(x) = \det(xI - T)\), the monic characteristic polynomial of \(T\).

    • \(V\) has exactly one invariant factor as a \(k[x]\)-module.

    • The rational canonical form of \(T\) has a single block.

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