Let \(V = \mathbb{C}^2\) with the standard inner product \(\langle u, v \rangle = u_1 \bar{v}_1 + u_2 \bar{v}_2\). Let \(T\) be the linear operator on \(V\) represented in the standard basis by the matrix: \[A = \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix}.\] (1) Prove that \(T\) is normal (\(A A^* = A^* A\)). (2) Find the eigenvalues of \(T\) and construct an orthonormal basis of \(\mathbb{C}^2\) consisting of eigenvectors of \(T\).
Solution
Goal: Prove normality of \(A\) and compute its spectral decomposition into an orthonormal eigenbasis.
1. Part 1: Verification of Normality:Proof:
1.1. The conjugate transpose (adjoint) of \(A\) is:\[A^* = \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix}^* = \begin{bmatrix} 1 & \bar{i} \\ \bar{i} & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & -i \\ -i & 1 \end{bmatrix}.\]
1.2. We compute the product \(A A^*\):\[A A^* = \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix} \begin{bmatrix} 1 & -i \\ -i & 1 \end{bmatrix} = \begin{bmatrix} 1 - i^2 & -i + i \\ i - i & -i^2 + 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2 I_2.\]
1.3. We compute the product \(A^* A\):\[A^* A = \begin{bmatrix} 1 & -i \\ -i & 1 \end{bmatrix} \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix} = \begin{bmatrix} 1 - i^2 & i - i \\ -i + i & -i^2 + 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2 I_2.\]
1.4. Since \(A A^* = A^* A = 2 I_2\), \(T\) is a normal operator.
2. Part 2: Eigenvalues of \(A\):Proof:
2.1. The characteristic polynomial of \(A\) is:\[p_A(\lambda) = \det(\lambda I - A) = \det \begin{bmatrix} \lambda - 1 & -i \\ -i & \lambda - 1 \end{bmatrix} = (\lambda - 1)^2 - (-i)^2 = (\lambda - 1)^2 - (-1) = (\lambda - 1)^2 + 1.\]
2.2. Setting \(p_A(\lambda) = 0\):\[(\lambda - 1)^2 = -1 \implies \lambda - 1 = \pm i \implies \lambda_1 = 1 + i, \quad \lambda_2 = 1 - i.\]
2.3. Since the two eigenvalues \(\lambda_1 \ne \lambda_2\) are distinct, their eigenspaces are orthogonal.
3. Part 3: Eigenvectors and Orthonormal Basis:Proof:
3.1. For \(\lambda_1 = 1 + i\):\[(A - (1+i)I) v = \begin{bmatrix} -i & i \\ i & -i \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies -i x + i y = 0 \implies x = y.\]An eigenvector is \(v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).Normalizing to unit length:\[u_1 = \frac{1}{\|v_1\|} v_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}.\]
3.2. For \(\lambda_2 = 1 - i\):\[(A - (1-i)I) v = \begin{bmatrix} i & i \\ i & i \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \implies i x + i y = 0 \implies x = -y.\]An eigenvector is \(v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).Normalizing to unit length:\[u_2 = \frac{1}{\|v_2\|} v_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}.\]
3.3. Inner product verification:\[\langle u_1, u_2 \rangle = \frac{1}{2} (1 \cdot 1 + 1 \cdot (-1)) = 0.\]
4. Conclusion:\(T\) is normal since \(A A^* = A^* A = 2 I_2\); an orthonormal eigenbasis is \(\mathcal{B} = \left\{ \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\}\) with eigenvalues \(1+i\) and \(1-i\). Q.E.D.