91Ó°ÊÓ

To find the eigenvalues of the Hermitian matrix \(\mathrm{H} = \left(\begin{array}{cc} 10 & 3i \ -3i & 2 \ \end{array}\right)\), solve the characteristic equation \(\det(\mathrm{H} - \lambda\mathrm{I}) = 0\).Calculating the determinant, you get: \(\left|\begin{array}{cc} 10 - \lambda & 3i \ -3i & 2 - \lambda \ \end{array}\right| = (10 - \lambda)(2 - \lambda) + 9 = \lambda^2 - 12\lambda + 29 = 0\).Solving for \(\lambda\), the eigenvalues are \(\lambda_1 = 6 + \sqrt{7}\) and \(\lambda_2 = 6 - \sqrt{7}\).

For each eigenvalue, solve \(\mathrm{H}\mathbf{v} = \lambda\mathbf{v}\) to find the corresponding eigenvector. Start with \(\lambda_1 = 6 + \sqrt{7}\):Set up the equation \(\left(\begin{array}{cc} 10 & 3i \ -3i & 2 \ \end{array}\right)\left(\begin{array}{c} v_1 \ v_2 \ \end{array}\right) = (6 + \sqrt{7})\left(\begin{array}{c} v_1 \ v_2 \ \end{array}\right)\).\Solves to \(v_2 = \frac{3i}{6 + \sqrt{7} - 10}v_1 = i v_1\). Thus, a normalized eigenvector is \(\mathbf{v}_1 = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \ i \ \end{array}\right)\). Now for \(\lambda_2 = 6 - \sqrt{7}\):Similarly solves to \(v_2 = \frac{3i}{6 - \sqrt{7} - 10}v_1 = -i v_1\). Thus, a normalized eigenvector is \(\mathbf{v}_2 = \frac{1}{\sqrt{2}}\left(\begin{array}{c} 1 \ -i \ \end{array}\right)\).

Use the normalized eigenvectors to form the columns of the unitary matrix \(\mathrm{U}\):\[ \mathrm{U} = \left(\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\ \ \frac{i}{\sqrt{2}} & -\frac{i}{\sqrt{2}} \end{array}\right). \]

Compute \(\mathrm{U}^{\dagger}\) (which is the conjugate transpose of \(\mathrm{U}\)) and verify that \(\mathrm{U}^{\dagger}\mathrm{HU} = \Lambda\), where \(\Lambda = \left(\begin{array}{cc} 6 + \sqrt{7} & 0\ 0 & 6 - \sqrt{7} \end{array}\right).\)