Let \(V\) be a finite-dimensional vector space over the field \(F\), and let \(T\) be a linear operator on \(V\). We ask when it is true that every non-zero vector in \(V\) is a cyclic vector for \(T\). Prove that this is the case if and only if the characteristic polynomial for \(T\) is irreducible over \(F\).
Solution
Write \(n=\dim V\) and let \(\chi_T\) be the characteristic polynomial.
1. If \(\chi_T\) is irreducible, every nonzero vector is cyclic.
Let \(0\ne\alpha\in V\), and let \(p_\alpha\) be its monic \(T\)-annihilator. The minimal polynomial \(m_T\) annihilates \(\alpha\), so \[ p_\alpha\mid m_T\mid\chi_T. \] The polynomial \(p_\alpha\) is nonconstant, since the constant polynomial \(1\) does not annihilate a nonzero vector. Because \(\chi_T\) is irreducible, the divisibility chain forces \[ p_\alpha=\chi_T. \] Thus \[ \dim Z(\alpha;T)=\deg p_\alpha=\deg\chi_T=n \] [@HK71]. Hence \(Z(\alpha;T)=V\), so \(\alpha\) is cyclic.
2. If every nonzero vector is cyclic, \(\chi_T\) is irreducible.
Choose any nonzero \(\alpha\). By hypothesis it is cyclic, so its annihilator is the minimal polynomial and has degree \(n\). Since the characteristic polynomial is also monic of degree \(n\) and is divisible by the minimal polynomial, \[ m_T=\chi_T. \]
Suppose \(\chi_T\) were reducible. Write \[ \chi_T=fg \] with monic polynomials \(f,g\) of positive degree strictly less than \(n\). Because the annihilator of the cyclic vector \(\alpha\) is exactly \(\chi_T\), \[ \beta=f(T)\alpha\ne0; \] otherwise the lower-degree polynomial \(f\) would annihilate \(\alpha\). But \[ g(T)\beta=(gf)(T)\alpha=\chi_T(T)\alpha=0. \] Thus the annihilator of \(\beta\) has degree at most \(\deg g<n\), so \[ \dim Z(\beta;T)<n. \] Therefore the nonzero vector \(\beta\) is not cyclic, contradicting the hypothesis. Hence \(\chi_T\) is irreducible.
3. Conclusion.
The two steps prove \[ \boxed{\text{every nonzero vector is cyclic} \iff \chi_T\text{ is irreducible over }F.} \]