Let \(T\) be the linear operator on \(C^3\) which is represented in the standard ordered basis by the matrix \[ \begin{bmatrix} 1 & i & 0 \\ -1 & 2 & -i \\ 0 & 1 & 1 \end{bmatrix}. \] Find the \(T\)-annihilator of the vector \((1, 0, 0)\). Find the \(T\)-annihilator of \((1, 0, i)\).
Solution
Goal: Determine the \(T\)-annihilator polynomials \(p_{v_1}(x)\) and \(p_{v_2}(x)\) for the vectors \(v_1 = (1, 0, 0)^T\) and \(v_2 = (1, 0, i)^T\) under the linear operator \(T\).
1. Definition of the \(T\)-annihilator:Proof:
1.1. For a linear operator \(T\) on a vector space \(V\) and a vector \(v \in V\), the \(T\)-annihilator of \(v\) (denoted \(p_v(x)\)) is the unique monic polynomial of minimal degree in \(F[x]\) such that \(p_v(T)v = 0\).
1.2. The degree of \(p_v(x)\) equals the dimension of the \(T\)-cyclic subspace \(Z(v; T) = \operatorname{span}\{v, Tv, T^2v, \dots\}\).
2. \(T\)-annihilator of \(v_1 = (1, 0, 0)^T\):Proof:
2.1. Compute the powers of \(T\) acting on \(v_1\):\[v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \qquad Tv_1 = \begin{bmatrix} 1 & i & 0 \\ -1 & 2 & -i \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix},\]\[T^2 v_1 = T(Tv_1) = \begin{bmatrix} 1 & i & 0 \\ -1 & 2 & -i \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 - i \\ -3 \\ -1 \end{bmatrix}.\]
2.2. Place the vectors \(v_1, Tv_1, T^2v_1\) as columns of a matrix and compute the determinant:\[\det \begin{bmatrix} 1 & 1 & 1-i \\ 0 & -1 & -3 \\ 0 & 0 & -1 \end{bmatrix} = 1 \cdot (-1) \cdot (-1) = 1 \neq 0.\]
2.3. Because the determinant is non-zero, the set \(\{v_1, Tv_1, T^2v_1\}\) is linearly independent in \(\mathbb{C}^3\).
2.4. Thus \(\dim Z(v_1; T) = 3 = \dim \mathbb{C}^3\), which implies \(\deg p_{v_1}(x) = 3\).
2.5. Consequently, \(p_{v_1}(x)\) must equal the characteristic polynomial \(\det(xI - T)\) of \(T\):\[\det(xI - T) = \det \begin{bmatrix} x - 1 & -i & 0 \\ 1 & x - 2 & i \\ 0 & -1 & x - 1 \end{bmatrix} = (x - 1)[(x - 2)(x - 1) + i] + i(x - 1) = (x - 1)(x^2 - 3x + 2 + 2i).\]
2.6. Expanding gives\[p_{v_1}(x) = x^3 - 4x^2 + (5 + 2i)x - (2 + 2i).\]
3. \(T\)-annihilator of \(v_2 = (1, 0, i)^T\):Proof:
3.1. Compute \(Tv_2\):\[Tv_2 = \begin{bmatrix} 1 & i & 0 \\ -1 & 2 & -i \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ i \end{bmatrix} = \begin{bmatrix} 1 + 0 + 0 \\ -1 + 0 - i^2 \\ 0 + 0 + i \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ i \end{bmatrix} = v_2.\]
3.2. Thus \((T - I)v_2 = 0\).
3.3. Since \(v_2 \neq 0\), the degree of the \(T\)-annihilator is at least 1, so the monic polynomial of degree 1 annihilating \(v_2\) is\[p_{v_2}(x) = x - 1.\]
4. Conclusion:Proof:The \(T\)-annihilator of \((1, 0, 0)\) is \(x^3 - 4x^2 + (5 + 2i)x - (2 + 2i)\), and the \(T\)-annihilator of \((1, 0, i)\) is \(x - 1\).