Let \(V\) be the space \(C^2\), with the standard inner product. Let \(T\) be the linear operator defined by \(T\epsilon_1 = (1, -2)\), \(T\epsilon_2 = (i, -1)\). If \(\alpha = (x_1, x_2)\), find \(T^*\alpha\).
Solution
1. Matrix representation of \(T\) in the standard orthonormal basis:
1.1. The standard basis \(\mathcal{B} = \{e_1, e_2\}\) is an orthonormal basis of \(\mathbb{C}^2\) under the standard inner product:\[ \langle (u_1, u_2), (v_1, v_2) \rangle = u_1 \overline{v_1} + u_2 \overline{v_2}. \]
1.2. From \(T e_1 = (1, -2) = 1 e_1 - 2 e_2\) and \(T e_2 = (i, -1) = i e_1 - 1 e_2\), the matrix of \(T\) with respect to \(\mathcal{B}\) is:\[ [T]_\mathcal{B} = \begin{pmatrix} 1 & i \\ -2 & -1 \end{pmatrix}. \]
2. Computation of the adjoint matrix \([T^*]_\mathcal{B}\):
2.1. With respect to an orthonormal basis, the matrix of the adjoint operator \(T^*\) is the conjugate transpose of \([T]_\mathcal{B}\):\[ [T^*]_\mathcal{B} = [T]_\mathcal{B}^* = \overline{[T]_\mathcal{B}}^T = \begin{pmatrix} 1 & \overline{-2} \\ \overline{i} & \overline{-1} \end{pmatrix}^T = \begin{pmatrix} 1 & -2 \\ -i & -1 \end{pmatrix}^T = \begin{pmatrix} 1 & -2 \\ -i & -1 \end{pmatrix}. \]
2.2. For any vector \(\alpha = (x_1, x_2) \in \mathbb{C}^2\):\[ [T^*\alpha]_\mathcal{B} = [T^*]_\mathcal{B} [\alpha]_\mathcal{B} = \begin{pmatrix} 1 & -2 \\ -i & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} x_1 - 2x_2 \\ -i x_1 - x_2 \end{pmatrix}. \]Thus:\[ T^*\alpha = (x_1 - 2x_2, \, -i x_1 - x_2). \]
3. Verification of the adjoint identity \(\langle T\beta, \alpha \rangle = \langle \beta, T^*\alpha \rangle\):
3.1. For \(\beta = (y_1, y_2)\):\[ \langle T\beta, \alpha \rangle = (y_1 + i y_2)\overline{x_1} + (-2y_1 - y_2)\overline{x_2} = y_1(\overline{x_1} - 2\overline{x_2}) + y_2(i \overline{x_1} - \overline{x_2}). \]\[ \langle \beta, T^*\alpha \rangle = y_1 \overline{(x_1 - 2x_2)} + y_2 \overline{(-i x_1 - x_2)} = y_1(\overline{x_1} - 2\overline{x_2}) + y_2(i \overline{x_1} - \overline{x_2}). \]Both expressions are equal.
4. Conclusion:\(T^*\alpha = (x_1 - 2x_2, -i x_1 - x_2)\). Q.E.D.