Let \(T\) be the linear operator on \(\RR^3\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \end{bmatrix}. \] Prove that \(T\) has no cyclic vector. What is the \(T\)-cyclic subspace generated by the vector \((1, -1, 3)\)?
Solution
Let \[ A=\operatorname{diag}(2,2,-1). \] The polynomial \[ q(x)=(x-2)(x+1)=x^2-x-2 \] annihilates \(T\), since \(q(2)=q(-1)=0\). Thus \[ T^2=T+2I. \] It follows inductively that every power \(T^k\) lies in the span of \(I\) and \(T\). Hence, for every vector \(v\), \[ Z(v;T)=\operatorname{span}\{v,Tv,T^2v,\ldots\} =\operatorname{span}\{v,Tv\}, \] so every cyclic subspace has dimension at most two. Since \(\dim\mathbb R^3=3\), no vector can be cyclic for \(T\) [@HK71].
For \[ v=(1,-1,3), \] one has \[ Tv=(2,-2,-3). \] These two vectors are linearly independent, because they are not scalar multiples. Therefore \[ \boxed{Z(v;T)=\operatorname{span}\{(1,-1,3),(2,-2,-3)\}.} \] Equivalently, adding and subtracting suitable multiples gives \[ Z(v;T)=\operatorname{span}\{(1,-1,0),(0,0,1)\}. \]