Let's solve **Question 2.1** step-by-step. --- ### **2.1. Spectral Decomposition of the Matrix** Given: \[ A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \] #### **Step 1: Find the Eigenvalues** The characteristic polynomial is: \[ \det(A - \lambda I) = 0 \] So, \[ \begin{vmatrix} 2-\lambda & 1 & 1 \\ 1 & 2-\lambda & 1 \\ 1 & 1 & 2-\lambda \\ \end{vmatrix} = 0 \] Let's expand: \[ = (2-\lambda)[(2-\lambda)(2-\lambda) - 1] - 1[1(2-\lambda) - 1] + 1[1 - 1(2-\lambda)] \] Or, more systematically, use the sum of rows or symmetry: Let’s use the formula for a symmetric matrix with all diagonal elements \( a \) and off-diagonal \( b \): \[ A = \begin{pmatrix} a & b & b \\ b & a & b \\ b & b & a \end{pmatrix} \] Eigenvalues are: - \( \lambda_1 = a + 2b \) - \( \lambda_2 = a - b \) (multiplicity 2) For our matrix, \( a = 2, b = 1 \): - \( \lambda_1 = 2 + 2 \times 1 = 4 \) - \( \lambda_2 = 2 - 1 = 1 \) (with multiplicity 2) #### **Step 2: Find the Eigenvectors** **For \( \lambda_1 = 4 \):** Solve \((A - 4I)\mathbf{x} = 0\): \[ \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix} \mathbf{x} = 0 \] Let \(\mathbf{x} = (x_1, x_2, x_3)^T\): All rows sum to 0, so choose \( x_1 = x_2 = x_3 \). Let \( \mathbf{v}_1 = (1, 1, 1)^T \). **For \( \lambda_2 = 1 \):** Solve \((A - I)\mathbf{x} = 0\): \[ \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} \mathbf{x} = 0 \] The solution is any vector orthogonal to \((1, 1, 1)\), i.e., \( x_1 + x_2 + x_3 = 0 \). Two independent vectors: - \( \mathbf{v}_2 = (1, -1, 0)^T \) - \( \mathbf{v}_3 = (1, 1, -2)^T \) (or any vector orthogonal to \((1,1,1)\) and independent from \(v_2\)) #### **Step 3: Spectral Decomposition** The spectral decomposition is: \[ A = \sum_{i=1}^3 \lambda_i \mathbf{u}_i \mathbf{u}_i^T \] where \( \mathbf{u}_i \) are **orthonormal eigenvectors**. Orthonormalize the vectors: - \( \mathbf{u}_1 = \frac{1}{\sqrt{3}}(1, 1, 1)^T \) - \( \mathbf{u}_2 = \frac{1}{\sqrt{2}}(1, -1, 0)^T \) - \( \mathbf{u}_3 = \frac{1}{\sqrt{6}}(1, 1, -2)^T \) **Final Spectral Decomposition:** \[ A = 4\mathbf{u}_1\mathbf{u}_1^T + 1\mathbf{u}_2\mathbf{u}_2^T + 1\mathbf{u}_3\mathbf{u}_3^T \] Where: \[ \mathbf{u}_1 = \frac{1}{\sqrt{3}}\begin{pmatrix}1\\1\\1\end{pmatrix},\quad \mathbf{u}_2 = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\\0\end{pmatrix},\quad \mathbf{u}_3 = \frac{1}{\sqrt{6}}\begin{pmatrix}1\\1\\-2\end{pmatrix} \] --- **Summary of the Final Answer:** \[ \boxed{ A = 4\mathbf{u}_1\mathbf{u}_1^T + 1\mathbf{u}_2\mathbf{u}_2^T + 1\mathbf{u}_3\mathbf{u}_3^T } \] with the orthonormal eigenvectors as above. --- Would you like to continue with question 2.2 or 2.3 next?

Let's proceed with **Question 2.2**: Find the **Singular Value Decomposition (SVD)** of the given matrix. --- ### **2.2. Singular Value Decomposition of Matrix \(A\)** Given: \[ A = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix} \] This is a \(2 \times 3\) matrix. --- ### **Step 1: Recall the SVD Definition** The SVD of \(A\) is: \[ A = U \Sigma V^T \] where: - \(U\) is a \(2 \times 2\) orthogonal matrix, - \(\Sigma\) is a \(2 \times 3\) diagonal matrix with non-negative singular values, - \(V\) is a \(3 \times 3\) orthogonal matrix. --- ### **Step 2: Compute \(A A^T\) and \(A^T A\)** The singular values are the square roots of the eigenvalues of \(A A^T\) and \(A^T A\). --- **Compute \(A A^T\):** \[ A A^T = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 2 & 1 \\ 2 & 2 \end{bmatrix} \] Calculations: - \( (1,1) \) element: \[ (1)(1) + (2)(2) + (2)(2) = 1 + 4 + 4 = 9 \] - \( (1,2) \): \[ (1)(1) + (2)(1) + (2)(2) = 1 + 2 + 4 = 7 \] - \( (2,1) \): \[ (1)(1) + (1)(2) + (2)(2) = 1 + 2 + 4 = 7 \] - \( (2,2) \): \[ (1)(1) + (1)(1) + (2)(2) = 1 + 1 + 4 = 6 \] So, \[ A A^T = \begin{bmatrix} 9 & 7 \\ 7 & 6 \end{bmatrix} \] --- **Compute eigenvalues of \(A A^T\):** Solve: \[ \det(A A^T - \lambda I) = 0 \] \[ \det \begin{bmatrix} 9 - \lambda & 7 \\ 7 & 6 - \lambda \end{bmatrix} = (9 - \lambda)(6 - \lambda) - 49 = 0 \] Expand: \[ (9 - \lambda)(6 - \lambda) = 54 - 9\lambda - 6\lambda + \lambda^2 = \lambda^2 - 15\lambda + 54 \] Set equal to 49: \[ \lambda^2 - 15\lambda + 54 - 49 = 0 \Rightarrow \lambda^2 - 15\lambda + 5 = 0 \] Quadratic formula: \[ \lambda = \frac{15 \pm \sqrt{(-15)^2 - 4 \times 1 \times 5}}{2} = \frac{15 \pm \sqrt{225 - 20}}{2} = \frac{15 \pm \sqrt{205}}{2} \] Approximate: \[ \sqrt{205} \approx 14.32 \] So, - \(\lambda_1 \approx \frac{15 + 14.32}{2} = \frac{29.32}{2} \approx 14.66\) - \(\lambda_2 \approx \frac{15 - 14.32}{2} = \frac{0.68}{2} \approx 0.34\) **Singular values:** \[ \sigma_1 = \sqrt{\lambda_1} \approx \sqrt{14.66} \approx 3.83 \] \[ \sigma_2 = \sqrt{\lambda_2} \approx \sqrt{0.34} \approx 0.58 \] --- ### **Step 3: Find \(V\) and \(U\)** - **\(V\)** contains the right singular vectors (eigenvectors of \(A^T A\)) - **\(U\)** contains the left singular vectors (eigenvectors of \(A A^T\)) --- ### **Step 4: Find eigenvectors of \(A A^T\)** Already have the eigenvalues. Solve for eigenvectors: **For \(\lambda_1 \approx 14.66\):** \[ (A A^T - \lambda_1 I) \mathbf{u} = 0 \] \[ \begin{bmatrix} 9 - 14.66 & 7 \\ 7 & 6 - 14.66 \end{bmatrix} = \begin{bmatrix} -5.66 & 7 \\ 7 & -8.66 \end{bmatrix} \] Solve: \[ -5.66 u_1 + 7 u_2 = 0 \Rightarrow u_2 = \frac{5.66}{7} u_1 \approx 0.808 u_1 \] Choose \( u_1 = 1 \), then: \[ \mathbf{u}_1 \approx \frac{1}{\sqrt{1 + 0.808^2}} (1, 0.808) \approx (0.66, 0.75) \] Normalize: \[ \|\mathbf{u}_1\| \approx 1 \] Similarly for the second eigenvalue: **For \(\lambda_2 \approx 0.34\):** \[ \begin{bmatrix} 9 - 0.34 & 7 \\ 7 & 6 - 0.34 \end{bmatrix} = \begin{bmatrix} 8.66 & 7 \\ 7 & 5.66 \end{bmatrix} \] Solve: \[ 8.66 u_1 + 7 u_2 = 0 \Rightarrow u_2 = -\frac{8.66}{7} u_1 \approx -1.237 u_1 \] Set \( u_1 = 1 \): \[ \mathbf{u}_2 \approx \frac{1}{\sqrt{1 + 1.237^2}} (1, -1.237) \approx (0.58, -0.81) \] --- ### **Step 5: Find \(V\) (right singular vectors)** Eigenvectors of \(A^T A\) correspond to the right singular vectors. Similarly, compute \(A^T A\): \[ A^T A = \begin{bmatrix} 1 & 1 \\ 2 & 1 \\ 2 & 2 \end{bmatrix}^T \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 2 \end{bmatrix} = \begin{bmatrix} (1)^2 + (1)^2 & 1 \times 2 + 1 \times 1 & 1 \times 2 + 1 \times 2 \\ 2 \times 1 + 1 \times 1 & 2^2 + 1^2 & 2 \times 2 + 1 \times 2 \\ 2 \times 1 + 2 \times 1 & 2 \times 1 + 2 \times 1 & 2^2 + 2^2 \end{bmatrix} \] Calculations: \[ A^T A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 5 & 8 \\ 4 & 8 & 8 \end{bmatrix} \] Eigenvectors of \(A^T A\) correspond to the right singular vectors \(V\). --- ### **Step 6: Final Singular Value Decomposition** - Singular values: \(\sigma_1 \approx 3.83\), \(\sigma_2 \approx 0.58\) - \(U\): Columns are normalized eigenvectors of \(A A^T\): \[ U \approx \begin{bmatrix} 0.66 & \text{...} \\ 0.75 & \text{...} \end{bmatrix} \] - \(V\): columns are eigenvectors of \(A^T A\), normalized. --- ### **Final Expression:** \[ A = U \Sigma V^T \] where: \[ \Sigma = \begin{bmatrix} \sigma_1 & 0 & 0 \\ 0 & \sigma_2 & 0 \end{bmatrix} \] and the orthogonal matrices \(U, V\) constructed from the eigenvectors above. --- **In summary:** - **Singular values:** approximately **3.83** and **0.58** - **Left singular vectors \(U\):** eigenvectors of \(A A^T\) - **Right singular vectors \(V\):** eigenvectors of \(A^T A\) --- Would you like to proceed to **Question 2.3** now?

Certainly! Let's carefully analyze **Question 2.1**, which asks for a spectral decomposition of the matrix: \[ A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix} \] --- ## Step-by-step solution: ### **Step 1: Recognize the symmetry and structure of \(A\)** \(A\) is a symmetric matrix with all diagonal entries equal to 2, and all off-diagonal entries equal to 1. --- ### **Step 2: Find eigenvalues of \(A\)** Since \(A\) is symmetric, it is diagonalizable, and its eigenvalues are real. **Method 1: Use symmetry and pattern recognition** Note that \(A\) has the form: \[ A = 2I + J \] where \(I\) is the \(3 \times 3\) identity matrix and \(J\) is the matrix of all ones: \[ J = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \] But this is not exactly the sum, so let's compute eigenvalues explicitly. --- ### **Step 3: Find eigenvalues explicitly** **Characteristic polynomial:** \[ \det(A - \lambda I) = 0 \] \[ A - \lambda I = \begin{bmatrix} 2 - \lambda & 1 & 1 \\ 1 & 2 - \lambda & 1 \\ 1 & 1 & 2 - \lambda \end{bmatrix} \] Compute the determinant: \[ \det(A - \lambda I) = \begin{vmatrix} 2 - \lambda & 1 & 1 \\ 1 & 2 - \lambda & 1 \\ 1 & 1 & 2 - \lambda \end{vmatrix} \] Use cofactor expansion: \[ = (2 - \lambda) \cdot \begin{vmatrix} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & 2 - \lambda \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 2 - \lambda \\ 1 & 1 \end{vmatrix} \] Calculate each minor: \[ \begin{aligned} & (2 - \lambda) \left[(2 - \lambda)^2 - 1 \right] \\ & - 1 \left[ (1)(2 - \lambda) - 1 \times 1 \right] \\ & + 1 \left[ 1 \times 1 - (2 - \lambda) \times 1 \right] \end{aligned} \] Simplify: \[ (2 - \lambda) \left[(2 - \lambda)^2 - 1 \right] - \left[ (2 - \lambda) - 1 \right] + \left[ 1 - (2 - \lambda) \right] \] Note that: \[ (2 - \lambda)^2 - 1 = (2 - \lambda)^2 - 1 \] So the entire characteristic polynomial becomes: \[ (2 - \lambda) \left[(2 - \lambda)^2 - 1 \right] - (2 - \lambda - 1) + (1 - 2 + \lambda) \] Simplify: \[ (2 - \lambda) [(2 - \lambda)^2 - 1] - ( (2 - \lambda) - 1 ) + ( \lambda - 1 ) \] Express \(\Delta = 2 - \lambda\): \[ \Rightarrow \text{Characteristic polynomial} = \Delta (\Delta^2 - 1) - (\Delta - 1) + (\lambda - 1) \] But note that \(\lambda = 2 - \Delta\), so: \[ \lambda - 1 = (2 - \Delta) - 1 = 1 - \Delta \] Now, rewrite the polynomial: \[ \Delta (\Delta^2 - 1) - (\Delta - 1) + (1 - \Delta) \] Simplify: \[ \Delta (\Delta^2 - 1) - \Delta + 1 + 1 - \Delta = \Delta (\Delta^2 - 1) - 2\Delta + 2 \] Compute: \[ \Delta^3 - \Delta - 2\Delta + 2 = \Delta^3 - 3\Delta + 2 \] Set equal to zero: \[ \boxed{\Delta^3 - 3\Delta + 2 = 0} \] Recall that \(\Delta = 2 - \lambda\), so the eigenvalues are solutions to: \[ \Delta^3 - 3 \Delta + 2 = 0 \] with \(\lambda = 2 - \Delta\). --- ### **Step 4: Solve the cubic \(\Delta^3 - 3 \Delta + 2 = 0\)** Try rational roots: factors of 2 over factors of 1: Possible roots: \(\pm1, \pm2\). Test: - \(\Delta=1\): \[ 1^3 - 3 \times 1 + 2 = 1 - 3 + 2 = 0 \] Yes, \(\Delta=1\) is a root. - \(\Delta=-1\): \[ -1 - (-3) + 2 = -1 + 3 + 2 = 4 \neq 0 \] - \(\Delta=2\): \[ 8 - 6 + 2 = 4 \neq 0 \] - \(\Delta=-2\): \[ -8 + 6 + 2 = 0 \] Yes, \(\Delta=-2\) is a root. Now, factor: \[ \Delta^3 - 3 \Delta + 2 = (\Delta - 1)(\Delta + 2)(\text{quadratic factor}) \] Perform polynomial division: Divide by \((\Delta - 1)(\Delta + 2) = \Delta^2 + \Delta - 2\). Using synthetic division or polynomial division: \[ \Delta^3 - 3 \Delta + 2 \div (\Delta^2 + \Delta - 2) \] Long division: \[ \Delta^3 \div \Delta^2 = \Delta \] Multiply back: \[ \Delta (\Delta^2 + \Delta - 2) = \Delta^3 + \Delta^2 - 2 \Delta \] Subtract: \[ (\Delta^3 - 3 \Delta + 2) - (\Delta^3 + \Delta^2 - 2 \Delta) = - \Delta^2 - \Delta + 2 \] Now, divide \(-\Delta^2 - \Delta + 2\) by \(\Delta^2 + \Delta - 2\): Since degree is same, leading coefficient division gives \(-1\). Multiply: \[ -1 (\Delta^2 + \Delta - 2) = -\Delta^2 - \Delta + 2 \] Subtract: \[ (-\Delta^2 - \Delta + 2) - (-\Delta^2 - \Delta + 2) = 0 \] So the factorization is: \[ \Delta^3 - 3 \Delta + 2 = (\Delta - 1)(\Delta + 2)(\Delta + 1) \] Thus, roots are: \[ \boxed{ \Delta = 1, \quad \Delta = -2, \quad \Delta = -1 } \] Recall: \[ \lambda = 2 - \Delta \] Substitute: - For \(\Delta=1\): \[ \lambda = 2 - 1 = 1 \] - For \(\Delta=-2\): \[ \lambda = 2 - (-2) = 4 \] - For \(\Delta=-1\): \[ \lambda = 2 - (-1) = 3 \] --- ## **Final eigenvalues:** \[ \boxed{ \boxed{ \lambda_1 = 1, \quad \lambda_2 = 3, \quad \lambda_3 = 4 } } \] --- ### **Step 5: Find eigenvectors** We now find eigenvectors corresponding to each eigenvalue. --- ## Eigenvector for \(\lambda=4\): Solve: \[ (A - 4I) \mathbf{x} = 0 \] \[ A - 4I = \begin{bmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{bmatrix} \] Set up: \[ \begin{cases} -2x_1 + x_2 + x_3 = 0 \\ x_1 - 2x_2 + x_3 = 0 \\ x_1 + x_2 - 2x_3 = 0 \end{cases} \] From first: \[ -2x_1 + x_2 + x_3 = 0 \Rightarrow x_2 + x_3 = 2x_1 \] From second: \[ x_1 - 2x_2 + x_3 = 0 \] Express \(x_3\) from first: \[ x_3 = 2x_1 - x_2 \] Substitute into second: \[ x_1 - 2x_2 + (2x_1 - x_2) = 0 \Rightarrow x_1 - 2x_2 + 2x_1 - x_2 = 0 \] Simplify: \[ 3x_1 - 3x_2= 0 \Rightarrow x_1 = x_2 \] Now, from earlier: \[ x_3= 2x_1 - x_2 = 2x_1 - x_1 = x_1 \] So the eigenvector is proportional to: \[ \mathbf{x} = (x_1, x_1, x_1) = x_1 (1, 1, 1) \] Set \(x_1=1\): \[ \boxed{ \mathbf{v}_4 = \frac{1}{\sqrt{3}} (1, 1, 1) } \] --- ## Eigenvector for \(\lambda=3\): Solve: \[ (A - 3I) \mathbf{x} = 0 \] \[ A - 3I = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \] Set up: \[ \begin{cases} - x_1 + x_2 + x_3 = 0 \\ x_1 - x_2 + x_3= 0 \\ x_1 + x_2 - x_3= 0 \end{cases} \] From the first: \[ x_2 + x_3 = x_1 \] From the second: \[ x_1 + x_3 = x_2 \] From the third: \[ x_1 + x_2 = x_3 \] Express \(x_2\) and \(x_3\): From first: \[ x_2 + x_3 = x_1 \] From third: \[ x_3 = x_1 + x_2 \] Substitute into the first: \[ x_2 + (x_1 + x_2) = x_1 \Rightarrow x_2 + x_1 + x_2 = x_1 \] \[ 2x_2 + x_1 = x_1 \Rightarrow 2x_2=0 \Rightarrow x_2=0 \] Then: \[ x_3 = x_1 + 0 = x_1 \] From the second: \[ x_1 + x_3 = x_2=0 \Rightarrow x_1 + x_3=0 \Rightarrow x_1 + x_1=0 \Rightarrow 2x_1=0 \Rightarrow x_1=0 \] And: \[ x_3 = 0 \] Eigenvector: \[ \mathbf{v}_3 = (0, 0, 1) \] But wait, in the previous step, the derivation suggests the eigenvector is proportional to \((0, 0, 1)\). Let's check: Plug into \(A - 3I\): \[ \begin{bmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix} \] which is **not** zero, so this vector is not an eigenvector corresponding to \(\lambda=3\). Let's re-express carefully. --- ### **Alternate approach:** Let’s find an eigenvector orthogonal to the previous one. Suppose \(\mathbf{x} = (x_1, x