Let's solve each part step-by-step as requested: --- ### (a) Prove that if λ is an eigenvalue of T, then λ is real. #### **Step 1:** Write the eigenvalue equation and use the definition of self-adjoint. Let \( T(v) = \lambda v \) for some nonzero \( v \in V \). Since \( T \) is self-adjoint, \( \langle T(v), v \rangle = \langle v, T(v) \rangle \). #### **Explanation:** The property of self-adjointness tells us how the operator interacts with the inner product. The eigenvalue equation gives us a relationship involving \( \lambda \). --- #### **Step 2:** Compute both sides using the eigenvalue equation. \[ \langle T(v), v \rangle = \langle \lambda v, v \rangle = \lambda \langle v, v \rangle \] \[ \langle v, T(v) \rangle = \langle v, \lambda v \rangle = \overline{\lambda} \langle v, v \rangle \] Since \( \langle T(v), v \rangle = \langle v, T(v) \rangle \), we have \( \lambda \langle v, v \rangle = \overline{\lambda} \langle v, v \rangle \). Because \( v \neq 0 \), \( \langle v, v \rangle \neq 0 \), so \( \lambda = \overline{\lambda} \), which means \( \lambda \) is real. #### **Explanation:** The equality of the two forms of the inner product, combined with the non-zero norm of \( v \), forces \( \lambda \) to be real. --- **Final Answer (a):** If \( \lambda \) is an eigenvalue of a self-adjoint operator \( T \), then \( \lambda \) is real. --- ### (b) Prove that eigenvectors corresponding to distinct eigenvalues are orthogonal. #### **Step 1:** Assume \( T(v_1) = \lambda_1 v_1 \) and \( T(v_2) = \lambda_2 v_2 \), \( \lambda_1 \neq \lambda_2 \). Consider \( \langle T(v_1), v_2 \rangle \) and \( \langle v_1, T(v_2) \rangle \). #### **Explanation:** We are considering two eigenvectors for different eigenvalues and will use the self-adjoint property. --- #### **Step 2:** Use the eigenvalue equations and self-adjointness. \[ \langle T(v_1), v_2 \rangle = \langle \lambda_1 v_1, v_2 \rangle = \lambda_1 \langle v_1, v_2 \rangle \] \[ \langle v_1, T(v_2) \rangle = \langle v_1, \lambda_2 v_2 \rangle = \overline{\lambda_2} \langle v_1, v_2 \rangle \] But since \( T \) is self-adjoint, \[ \langle T(v_1), v_2 \rangle = \langle v_1, T(v_2) \rangle \] So, \[ \lambda_1 \langle v_1, v_2 \rangle = \lambda_2 \langle v_1, v_2 \rangle \] Thus, \( (\lambda_1 - \lambda_2) \langle v_1, v_2 \rangle = 0 \). Since \( \lambda_1 \neq \lambda_2 \), \( \langle v_1, v_2 \rangle = 0 \). #### **Explanation:** The only way the above equation holds for distinct eigenvalues is if the inner product is zero, i.e., the vectors are orthogonal. --- **Final Answer (b):** Eigenvectors corresponding to distinct eigenvalues of a self-adjoint operator are orthogonal. --- ### (c) Example: 2×2 real self-adjoint (symmetric) matrix, find its eigenvalues and eigenvectors. #### **Step 1:** Choose a symmetric matrix. Example: \( A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \) Find eigenvalues by solving \( \det(A - \lambda I) = 0 \): \[ \begin{vmatrix} 2-\lambda & 1 \\ 1 & 3-\lambda \end{vmatrix} = (2-\lambda)(3-\lambda) - 1 = \lambda^2 - 5\lambda + 5 = 0 \] \[ \lambda = \frac{5 \pm \sqrt{25-20}}{2} = \frac{5 \pm \sqrt{5}}{2} \] #### **Explanation:** The matrix is symmetric (self-adjoint), and we have found its (real) eigenvalues. --- #### **Step 2:** Find eigenvectors for each eigenvalue. Let’s use \( \lambda_1 = \frac{5 + \sqrt{5}}{2} \): \[ (A - \lambda_1 I)x = 0 \Rightarrow \begin{pmatrix} 2-\lambda_1 & 1 \\ 1 & 3-\lambda_1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 \] Let’s solve for \( x_2 \) in terms of \( x_1 \) using the first row: \[ (2-\lambda_1)x_1 + x_2 = 0 \implies x_2 = (\lambda_1 - 2)x_1 \] So, the eigenvector is proportional to \( \begin{pmatrix} 1 \\ \lambda_1 - 2 \end{pmatrix} \). Similarly, for \( \lambda_2 = \frac{5 - \sqrt{5}}{2} \): \[ x_2 = (\lambda_2 - 2)x_1 \] Eigenvector is proportional to \( \begin{pmatrix} 1 \\ \lambda_2 - 2 \end{pmatrix} \). #### **Explanation:** We’ve constructed all eigenvalues and eigenvectors for the example symmetric (self-adjoint) matrix. --- **Final Answer (c):** Example matrix: \( \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \) Eigenvalues: \( \frac{5 \pm \sqrt{5}}{2} \) Eigenvectors: \( \begin{pmatrix} 1 \\ \frac{3+\sqrt{5}}{2} \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ \frac{3-\sqrt{5}}{2} \end{pmatrix} \) (up to scalar multiple).

**(a) Proof that eigenvalues of a self-adjoint operator are real:** **Step 1:** Suppose \( v \neq 0 \) is an eigenvector of \( T \) with eigenvalue \( \lambda \), so \( T(v) = \lambda v \). Using the property of self-adjointness, for this \( v \): \[ \langle T(v), v \rangle = \langle v, T(v) \rangle \] which simplifies to: \[ \langle \lambda v, v \rangle = \langle v, \lambda v \rangle \] **Explanation:** The inner product with a scalar multiple factors out the scalar, so this relates the eigenvalue \( \lambda \) to the inner product. --- **Step 2:** Calculate both sides: \[ \lambda \langle v, v \rangle = \overline{\lambda} \langle v, v \rangle \] Since \( v \neq 0 \), \( \langle v, v \rangle \neq 0 \), so dividing both sides by \( \langle v, v \rangle \): \[ \lambda = \overline{\lambda} \] which means \( \lambda \) is real. **Final answer:** **Eigenvalues of a self-adjoint operator are real.** --- **(b) Proof that eigenvectors corresponding to distinct eigenvalues are orthogonal:** **Step 1:** Let \( v_1, v_2 \) be eigenvectors with eigenvalues \( \lambda_1, \lambda_2 \), respectively, with \( \lambda_1 \neq \lambda_2 \): \[ T v_1 = \lambda_1 v_1,\quad T v_2 = \lambda_2 v_2 \] Using the self-adjoint property: \[ \langle T v_1, v_2 \rangle = \langle v_1, T v_2 \rangle \] **Explanation:** This step sets up the key relation that connects the inner products involving the operator \( T \). --- **Step 2:** Substitute the eigenvalue equations: \[ \langle \lambda_1 v_1, v_2 \rangle = \langle v_1, \lambda_2 v_2 \rangle \] which simplifies to: \[ \lambda_1 \langle v_1, v_2 \rangle = \overline{\lambda_2} \langle v_1, v_2 \rangle \] Since eigenvalues are real (from part (a)), \( \overline{\lambda_2} = \lambda_2 \). Thus: \[ (\lambda_1 - \lambda_2) \langle v_1, v_2 \rangle = 0 \] Because \( \lambda_1 \neq \lambda_2 \), it follows: \[ \boxed{\langle v_1, v_2 \rangle = 0} \] so the eigenvectors are orthogonal. **Final answer:** **Eigenvectors corresponding to different eigenvalues of a self-adjoint operator are orthogonal.** --- **(c) Example:** Consider the symmetric (self-adjoint) matrix: \[ A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \] **Step 1:** Find eigenvalues by solving \( \det(A - \lambda I) = 0 \): \[ \det \begin{pmatrix} 2 - \lambda & 1 \\ 1 & 3 - \lambda \end{pmatrix} = (2 - \lambda)(3 - \lambda) - 1 = 0 \] \[ (2 - \lambda)(3 - \lambda) = 1 \] \[ (6 - 2\lambda - 3\lambda + \lambda^2) = 1 \] \[ \lambda^2 - 5 \lambda + 6 = 1 \] \[ \lambda^2 - 5 \lambda + 5 = 0 \] Solve: \[ \lambda = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2} \] **Step 2:** Find eigenvectors. For \( \lambda_1 = \frac{5 + \sqrt{5}}{2} \): \[ (A - \lambda_1 I) v = 0 \Rightarrow \begin{pmatrix} 2 - \lambda_1 & 1 \\ 1 & 3 - \lambda_1 \end{pmatrix} v = 0 \] Calculate: \[ 2 - \lambda_1 = 2 - \frac{5 + \sqrt{5}}{2} = \frac{4 - 5 - \sqrt{5}}{2} = \frac{-1 - \sqrt{5}}{2} \] \[ 3 - \lambda_1 = 3 - \frac{5 + \sqrt{5}}{2} = \frac{6 - 5 - \sqrt{5}}{2} = \frac{1 - \sqrt{5}}{2} \] Eigenvector \( v^{(1)} \) satisfies: \[ \left( \frac{-1 - \sqrt{5}}{2} \right) x + y = 0 \Rightarrow y = \left( \frac{1 + \sqrt{5}}{2} \right) x \] Choosing \( x=1 \): \[ v^{(1)} = \begin{pmatrix} 1 \\ \frac{1 + \sqrt{5}}{2} \end{pmatrix} \] Similarly, for \( \lambda_2 = \frac{5 - \sqrt{5}}{2} \): \[ 2 - \lambda_2 = \frac{-1 + \sqrt{5}}{2} \] \[ 3 - \lambda_2 = \frac{1 + \sqrt{5}}{2} \] Eigenvector \( v^{(2)} \): \[ \left( \frac{-1 + \sqrt{5}}{2} \right) x + y= 0 \Rightarrow y = \left( \frac{1 - \sqrt{5}}{2} \right) x \] Choose \( x=1 \): \[ v^{(2)} = \begin{pmatrix} 1 \\ \frac{1 - \sqrt{5}}{2} \end{pmatrix} \] **Final answer:** An example symmetric matrix: \[ A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \] has eigenvalues: \[ \lambda_{1,2} = \frac{5 \pm \sqrt{5}}{2} \] with corresponding eigenvectors proportional to: \[ \begin{pmatrix} 1 \\ \frac{1 + \sqrt{5}}{2} \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 \\ \frac{1 - \sqrt{5}}{2} \end{pmatrix} \] --- **In summary:** **A:** Eigenvalues are real. **B:** Eigenvectors for distinct eigenvalues are orthogonal. **C:** A concrete symmetric matrix example with explicit eigenvalues and eigenvectors.