16.-Linear-Algebra-by-Kenneth-Hoffman · Problem 7.1.8

The Rational and Jordan Forms

Mathematics — Linear Algebra · cyclic-vectors-commuting-operators

Problem

Let TT be a linear operator on the finite-dimensional vector space VV. Suppose TT has a cyclic vector. Prove that if UU is any linear operator which commutes with TT, then UU is a polynomial in TT.
Intuitive explanation

A cyclic vector vv means the orbit {v,Tv,T2v,…}\{v, Tv, T^2v, \ldots\} spans all of VV—so TT is completely determined by how it acts on this single vector. The key insight is that any operator UU commuting with TT must respect this cyclic structure: if UU and TT commute, then UU preserves the span of powers of TT applied to vv. This forces UU to act on the basis {v,Tv,T2v,…}\{v, Tv, T^2v, \ldots\} in a way that's entirely determined by where UU sends vv, which must itself be a polynomial combination of these basis vectors.

Intuitive explanation

A cyclic vector vv means the orbit {v,Tv,T2v,…}\{v, Tv, T^2v, \ldots\} spans all of VV—so TT is completely determined by how it acts on this single vector. The key insight is that any operator UU commuting with TT must respect this cyclic structure: if UU and TT commute, then UU preserves the span of powers of TT applied to vv. This forces UU to act on the basis {v,Tv,T2v,…}\{v, Tv, T^2v, \ldots\} in a way that's entirely determined by where UU sends vv, which must itself be a polynomial combination of these basis vectors.

Related theory

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