91Ó°ÊÓ

The given matrix A has a Schur decomposition, which means it can be written as: \[ A = U T U^{H} \] where U is a unitary matrix and T is an upper triangular matrix. Since A is a 2x2 matrix, we have: \[ T = \begin{pmatrix} t_{11} & t_{12} \\ 0 & t_{22} \end{pmatrix}\] Additionally, we know that \(t_{12} \neq 0\).

To find the eigenvalues of A, we look at the characteristic equation. For a general 2x2 matrix: \[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\] The characteristic equation is given by: \[ det(A - \lambda I) = 0 \] In our case, we have: \[ det(T - \lambda I) = \begin{vmatrix} t_{11} - \lambda & t_{12} \\ 0 & t_{22} - \lambda \end{vmatrix} = (t_{11} - \lambda)(t_{22} - \lambda) = 0\] Thus, the eigenvalues of A are: \[ \lambda_1 = t_{11} \text{ and } \lambda_2 = t_{22} \]

We now show that \(\mathbf{u}_1\) is an eigenvector of A belonging to \(\lambda_1=t_{11}\) and \(\mathbf{u}_2\) is not an eigenvector of A belonging to \(\lambda_2=t_{22}\). (a) Since U is unitary, its columns are orthonormal, and we can denote the first column as \(\mathbf{u}_1\). For A to have an eigenvector corresponding to \(\lambda_1=t_{11}\), we must have: \[ A \mathbf{u}_1 = \lambda_1 \mathbf{u}_1 \] Using the Schur decomposition of A, we get: \[ U T U^{H} \mathbf{u}_1 = t_{11} \mathbf{u}_1 \] Since \(\mathbf{u}_1\) is the first column of U, we have \(U^{H} \mathbf{u}_1 = \mathbf{e}_1\), where \(\mathbf{e}_1\) is the first standard basis vector. Therefore, \[ U T \mathbf{e}_1 = t_{11} \mathbf{u}_1 \] Because T is an upper triangular matrix, we have \(T \mathbf{e}_1 = \begin{pmatrix} t_{11} \\ 0 \end{pmatrix}\), so: \[ U \begin{pmatrix} t_{11} \\ 0 \end{pmatrix} = t_{11} \mathbf{u}_1 \] Since the first entry of \(\mathbf{u}_1\) is non-zero (otherwise U wouldn't be unitary), this equation implies that \(\mathbf{u}_1\) is indeed an eigenvector of A belonging to \(\lambda_1=t_{11}\). (b) Now, let's check if \(\mathbf{u}_2\) is an eigenvector of A belonging to \(\lambda_2=t_{22}\). We can start with: \[ A \mathbf{u}_2 = \lambda_2 \mathbf{u}_2 \] Following a similar process as before: \[ U T U^{H} \mathbf{u}_2 = t_{22} \mathbf{u}_2 \] Since \(\mathbf{u}_2\) is the second column of U, we have \(U^{H} \mathbf{u}_2 = \mathbf{e}_2\), where \(\mathbf{e}_2\) is the second standard basis vector. Therefore, \[ U T \mathbf{e}_2 = t_{22} \mathbf{u}_2 \] Because T is an upper triangular matrix, we have \(T \mathbf{e}_2 = \begin{pmatrix} t_{12} \\ t_{22} \end{pmatrix}\), so: \[ U \begin{pmatrix} t_{12} \\ t_{22} \end{pmatrix} = t_{22} \mathbf{u}_2 \] Since \(t_{12} \neq 0\), this equation implies that \(\mathbf{u}_2\) indeed is not an eigenvector of A belonging to \(\lambda_2=t_{22}\). In conclusion, we have shown that the eigenvalues of A are \(\lambda_1 = t_{11}\) and \(\lambda_2 = t_{22}\), with \(\mathbf{u}_1\) being an eigenvector corresponding to \(\lambda_1\) and \(\mathbf{u}_2\) not being an eigenvector corresponding to \(\lambda_2\).