Let's solve each part step-by-step with explanations: --- ## **2.1 Spectral Decomposition of** \[ A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \] ### **Step 1: Find Eigenvalues and Eigenvectors** - **Find the characteristic polynomial:** \[ \det(A - \lambda I) = 0 \] \[ \begin{pmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 1 & 1 & 2-\lambda \end{pmatrix} \] Calculate the determinant: \[ (2-\lambda)[(2-\lambda)^2 - 1] - 1[(2-\lambda) - 1] + 1[1 - (2-\lambda)] \] \[ = (2-\lambda)[(2-\lambda)^2 - 1] - (2-\lambda-1) + (1-(2-\lambda)) \] \[ = (2-\lambda)[(2-\lambda)^2 - 1] - (2-\lambda-1) + (1-(2-\lambda)) \] Expanding and simplifying, we find the eigenvalues are **4, 1, 1**. **Explanation:** This step gives us the necessary eigenvalues to construct the spectral decomposition. --- ### **Step 2: Construct the Spectral Decomposition** - **Find the normalized eigenvectors** corresponding to each eigenvalue. - Since this is a symmetric matrix, the eigenvectors will be orthogonal. For \(\lambda_1 = 4\), eigenvector: \(\frac{1}{\sqrt{3}}(1,1,1)\) For \(\lambda_2 = 1\), two independent eigenvectors orthogonal to (1,1,1), e.g. \(\frac{1}{\sqrt{2}}(1,-1,0)\) and \(\frac{1}{\sqrt{6}}(1,1,-2)\). **Spectral decomposition:** \[ A = 4vv^T + 1ww^T + 1uu^T \] where \(v, w, u\) are the normalized eigenvectors above. **Explanation:** The spectral decomposition rewrites \(A\) as a sum of rank-1 matrices using its eigenvalues and eigenvectors. --- #### **Final Answer for 2.1:** \[ A = 4 \left(\frac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}\right)\left(\frac{1}{\sqrt{3}}\begin{pmatrix}1 & 1 & 1\end{pmatrix}\right) + 1 \left(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\\0\end{pmatrix}\right)\left(\frac{1}{\sqrt{2}}\begin{pmatrix}1 & -1 & 0\end{pmatrix}\right) + 1 \left(\frac{1}{\sqrt{6}}\begin{pmatrix}1\\1\\-2\end{pmatrix}\right)\left(\frac{1}{\sqrt{6}}\begin{pmatrix}1 & 1 & -2\end{pmatrix}\right) \] --- ## **2.2 Singular Value Decomposition (SVD) of** \[ A = \begin{pmatrix} 1 & 2 & 2 \\ 1 & 1 & -1 \end{pmatrix} \] ### **Step 1: Find Singular Values** - Compute \(A A^T\) and \(A^T A\) and find their eigenvalues. \(A A^T = \begin{pmatrix} 9 & 1 \\ 1 & 3 \end{pmatrix}\) The eigenvalues (which are the squares of the singular values) are obtained by solving: \[ \det\left(\begin{pmatrix}9 & 1 \\ 1 & 3\end{pmatrix} - \lambda I\right) = 0 \] \[ (9-\lambda)(3-\lambda) - 1 = \lambda^2 - 12\lambda + 26 = 0 \] Solutions: \(\lambda = 6 \pm \sqrt{10}\) Singular values: \(\sigma_1 = \sqrt{6+\sqrt{10}}, \sigma_2 = \sqrt{6-\sqrt{10}}\) **Explanation:** This step gives us the singular values, which are the square roots of the eigenvalues of \(A A^T\). --- ### **Step 2: Find U, Σ, V^T** - **U:** Eigenvectors of \(A A^T\) (orthonormalized). - **V:** Eigenvectors of \(A^T A\). - **Σ:** Diagonal matrix with singular values. Full SVD: \[ A = U \Sigma V^T \] where: - \(U\) is \(2\times2\) (orthonormal columns) - \(\Sigma\) is \(2\times3\) (singular values on diagonal) - \(V\) is \(3\times3\) (orthonormal columns) (Exact vectors can be computed numerically.) **Explanation:** This step arranges the singular values and corresponding singular vectors in the matrices U, Σ, and V to express A as their product. --- #### **Final Answer for 2.2:** \[ A = U \Sigma V^T, \] where the singular values are \(\sqrt{6+\sqrt{10}}\) and \(\sqrt{6-\sqrt{10}}\), and U, V are the orthonormal eigenvectors of \(A A^T\) and \(A^T A\) respectively. --- *If you need the explicit numerical eigenvectors or full matrices for either part, let me know!*