Let's break down the first part of the problem step-by-step as requested. We'll work on **1(a)** and **1(b)**, with explanations after each step. --- ### **Step 1: Find the SVD of \( A \) by hand** #### **Matrix \( A \):** \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 8 & 12 \\ 1 & 2 & 3 \\ 2 & 4 & 6 \\ -1 & -2 & -3 \end{bmatrix} \] #### **Explanation:** - SVD decomposes \( A \) as \( A = U \Sigma V^T \). - \( U \) is \( 5 \times 5 \) orthogonal, \( \Sigma \) is \( 5 \times 3 \) diagonal, \( V \) is \( 3 \times 3 \) orthogonal. #### **Observations:** - All rows of \( A \) are multiples of \( [1\ 2\ 3] \), so \( \text{rank}(A) = 1 \). - That means only the first singular value is nonzero. #### **Compute the largest singular value (\( \sigma_1 \)):** - The nonzero singular value is \( \sqrt{\text{sum of squares of all elements of \( A \)}} \), since all rows are proportional. - Compute the norm of the first column vector \( v = [1\ 4\ 1\ 2\ -1]^T \). But since we're dealing with row multiples, let's compute the norm of the full matrix: \[ \|A\|_F^2 = \sum_{i,j} A_{ij}^2 \] \[ \text{First row: } (1^2 + 2^2 + 3^2) = 1+4+9 = 14 \] \[ \text{Second row: } (4^2 + 8^2 + 12^2) = 16+64+144 = 224 \] \[ \text{Third row: } (1^2 + 2^2 + 3^2) = 14 \] \[ \text{Fourth row: } (2^2 + 4^2 + 6^2) = 4+16+36 = 56 \] \[ \text{Fifth row: } ((-1)^2 + (-2)^2 + (-3)^2) = 1+4+9=14 \] \[ \text{Total: } 14+224+14+56+14 = 322 \] So the Frobenius norm is \( \sqrt{322} \), and since there's only one nonzero singular value: \[ \sigma_1 = \sqrt{322} \] #### **Constructing \( U \), \( \Sigma \), \( V \):** - \( V \): The direction of the row vector, normalized: \( v = \frac{1}{\sqrt{14}}[1\ 2\ 3]^T \) - \( U \): The direction of the column vector, normalized: \( u = \frac{1}{\sqrt{66}}[1\ 4\ 1\ 2\ -1]^T \) - \( \Sigma \): Only \( \sigma_1 \) nonzero. So, \[ A = \left(\frac{1}{\sqrt{66}}[1\ 4\ 1\ 2\ -1] \right) \cdot \sqrt{322} \cdot \left(\frac{1}{\sqrt{14}}[1\ 2\ 3]^T\right) \] **Step 1 Explanation:** Since all rows of \( A \) are multiples of the vector \( [1\ 2\ 3] \), the matrix is rank 1. Thus, its SVD consists of one nonzero singular value, and the corresponding singular vectors are proportional to the all-rows and all-columns direction. --- ### **Step 2: Find the Moore-Penrose Pseudoinverse \( A^{\dagger} \) by hand** For a rank-1 matrix: \[ A = \sigma_1 u_1 v_1^T \] \[ A^{\dagger} = v_1 \frac{1}{\sigma_1} u_1^T \] From above: - \( u_1 = \frac{1}{\sqrt{66}}[1\ 4\ 1\ 2\ -1]^T \) - \( v_1 = \frac{1}{\sqrt{14}}[1\ 2\ 3]^T \) - \( \sigma_1 = \sqrt{322} \) \[ A^{\dagger} = v_1 \frac{1}{\sigma_1} u_1^T \] \[ A^{\dagger} = \left(\frac{1}{\sqrt{14}} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\right) \cdot \frac{1}{\sqrt{322}} \cdot \left(\frac{1}{\sqrt{66}} \begin{bmatrix} 1 & 4 & 1 & 2 & -1 \end{bmatrix}\right) \] \[ A^{\dagger}_{3 \times 5} = \frac{1}{\sqrt{14} \cdot \sqrt{322} \cdot \sqrt{66}} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 1 & 4 & 1 & 2 & -1 \end{bmatrix} \] **Step 2 Explanation:** For a rank-1 matrix, the pseudoinverse is constructed using the singular vectors and the inverse of the nonzero singular value. --- ## **Final Answer for Steps 1 and 2** **Step 1:** The SVD of \( A \) is: \[ A = U \Sigma V^T \] where \( U \) and \( V \) are as above, and \(\Sigma\) has only one nonzero value, \(\sqrt{322}\). **Step 2:** The pseudoinverse is: \[ A^{\dagger} = v_1 \frac{1}{\sigma_1} u_1^T \] with explicit form as shown above. --- **Let me know if you'd like to proceed with step (c) next!**