Let \(T\) be the linear operator on \(F^2\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}. \] Let \(\alpha_1 = (0, 1)\). Show that \(F^2 \neq Z(\alpha_1; T)\) and that there is no non-zero vector \(\alpha_2\) in \(F^2\) with \(Z(\alpha_2; T)\) disjoint from \(Z(\alpha_1; T)\).
Solution
Let \(e_1=(1,0)\) and \(e_2=(0,1)\). The given matrix satisfies \[ T(x,y)=(0,x), \] so \(Te_1=e_2\) and \(Te_2=0\).
1. The cyclic subspace generated by \(\alpha_1=e_2\) is proper.
Since \(T\alpha_1=0\), every polynomial in \(T\) applied to \(\alpha_1\) is a scalar multiple of \(\alpha_1\). Hence \[ Z(\alpha_1;T)=\operatorname{span}\{e_2\}. \] This is one-dimensional, whereas \(F^2\) is two-dimensional, so \[ F^2\ne Z(\alpha_1;T). \]
2. Every nonzero cyclic subspace meets \(Z(\alpha_1;T)\) nontrivially.
Let \(\alpha_2=(a,b)\ne0\).
If \(a=0\), then \(\alpha_2=be_2\) with \(b\ne0\), so \[ Z(\alpha_2;T)=\operatorname{span}\{e_2\}=Z(\alpha_1;T). \] Their intersection is therefore nonzero.
If \(a\ne0\), then \[ T\alpha_2=(0,a)=ae_2\ne0. \] Thus \(e_2\in Z(\alpha_2;T)\). Since \(e_2\in Z(\alpha_1;T)\) as well, \[ Z(\alpha_2;T)\cap Z(\alpha_1;T) \] contains the nonzero vector \(e_2\). In fact, \(\alpha_2\) and \(T\alpha_2\) are linearly independent when \(a\ne0\), so then \(Z(\alpha_2;T)=F^2\).
Hence no nonzero \(\alpha_2\) generates a cyclic subspace disjoint from \(Z(\alpha_1;T)\) [@HK71].