Let \(T\) be a linear operator on a finite-dimensional vector space and let \(A\) be the matrix of \(T\) in some ordered basis. Then \(T\) has a cyclic vector if and only if the determinants of the \((n-1) \times (n-1)\) submatrices of \(xI - A\) are relatively prime.
Solution
Let \[ f_1\mid f_2\mid\cdots\mid f_n \] be the invariant factors of the polynomial matrix \(xI-A\), with units included when necessary. The Smith-normal-form determinantal-divisor theorem gives \[ \Delta_{n-1}:=\gcd\{\det M:M\text{ an }(n-1)\times(n-1) \text{ submatrix of }xI-A\} =f_1\cdots f_{n-1} \] up to a nonzero scalar [@HK71]. Therefore the indicated minors are relatively prime exactly when \[ f_1\cdots f_{n-1}=1. \] Because the \(f_i\) are monic and satisfy the divisibility chain, this is equivalent to \[ f_1=\cdots=f_{n-1}=1. \]
By the rational canonical form, the nonunit invariant factors are precisely the annihilators of the cyclic summands in the cyclic decomposition of \(V\) under \(T\) [@HK71]. Thus \(T\) has a cyclic vector exactly when there is a single nonunit invariant factor, necessarily \(f_n\); equivalently, exactly when \(f_1=\cdots=f_{n-1}=1\). Combining the two equivalences proves the claim.