Let \(F\) be a subfield of the complex numbers and let \(T\) be the linear operator on \(F^4\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & a & 2 & 0 \\ 0 & 0 & b & 2 \end{bmatrix}. \] Find the characteristic polynomial for \(T\). Consider the cases \(a = b = 1\); \(a = b = 0\); \(a = 0\), \(b = 1\). In each of these cases, find the minimal polynomial for \(T\) and non-zero vectors \(\alpha_1, \ldots, \alpha_r\) which satisfy the conditions of Theorem 3.
Solution
Write \[ N=T-2I. \] For the standard basis \(e_1,e_2,e_3,e_4\), \[ Ne_1=e_2, \qquad Ne_2=ae_3, \qquad Ne_3=be_4, \qquad Ne_4=0. \] Because the matrix of \(T\) is triangular with diagonal entries all equal to \(2\), \[ \boxed{\chi_T(x)=(x-2)^4} \] in every case [@HK71].
Theorem 3 asks for nonzero vectors \(\alpha_1,\ldots,\alpha_r\) such that \[ F^4=Z(\alpha_1;T)\oplus\cdots\oplus Z(\alpha_r;T) \] and, if \(p_i\) is the \(T\)-annihilator of \(\alpha_i\), then \(p_{i+1}\mid p_i\) [@HK71].
1. Case \(a=b=1\).
Here \[ Ne_1=e_2, \quad N^2e_1=e_3, \quad N^3e_1=e_4, \quad N^4e_1=0. \] Thus \[ Z(e_1;T)=\operatorname{span}\{e_1,e_2,e_3,e_4\}=F^4. \] Since \(N^3e_1\ne0\) but \(N^4e_1=0\), the \(T\)-annihilator of \(e_1\) is \((x-2)^4\). Hence \[ \boxed{m_T(x)=(x-2)^4,\qquad r=1,\qquad \alpha_1=e_1.} \]
2. Case \(a=b=0\).
Now \[ Ne_1=e_2, \qquad Ne_2=Ne_3=Ne_4=0. \] Therefore \(N^2=0\) but \(N\ne0\), so \[ \boxed{m_T(x)=(x-2)^2.} \] Take \[ \alpha_1=e_1, \qquad \alpha_2=e_3, \qquad \alpha_3=e_4. \] Then \[ Z(e_1;T)=\operatorname{span}\{e_1,e_2\}, \qquad Z(e_3;T)=\operatorname{span}\{e_3\}, \qquad Z(e_4;T)=\operatorname{span}\{e_4\}. \] These subspaces form a direct sum equal to \(F^4\). Their annihilators are \[ p_1=(x-2)^2, \qquad p_2=p_3=x-2, \] and indeed \(p_2\mid p_1\) and \(p_3\mid p_2\).
3. Case \(a=0\), \(b=1\).
Here \[ Ne_1=e_2, \qquad Ne_2=0, \qquad Ne_3=e_4, \qquad Ne_4=0. \] Again \(N^2=0\) and \(N\ne0\), so \[ \boxed{m_T(x)=(x-2)^2.} \] Choose \[ \alpha_1=e_1, \qquad \alpha_2=e_3. \] Then \[ Z(e_1;T)=\operatorname{span}\{e_1,e_2\}, \qquad Z(e_3;T)=\operatorname{span}\{e_3,e_4\}, \] so \[ F^4=Z(e_1;T)\oplus Z(e_3;T). \] Both vectors have \(T\)-annihilator \((x-2)^2\), so the divisibility condition is also satisfied. Hence one may take \[ \boxed{r=2,\qquad \alpha_1=e_1,\ \alpha_2=e_3.} \]