Let's carefully work through this problem step by step. --- ## **Given Data** - Standard basis vectors: \[ e_1 = (1,0,0), \quad e_2 = (0,1,0), \quad e_3 = (0,0,1) \] - Operator \( M \) acts on \( e \) as: \[ M e = \left[ \begin{bmatrix} 1 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & -1 & 1 \end{bmatrix} \right] e \] - Vector \( v_e = (1,2,3) \) in standard basis. - Non-orthonormal basis \( b = \{b_1, b_2, b_3\} \): \[ b_1 = (1,1,0), \quad b_2 = (0,1,1), \quad b_3 = (1,0,1) \] --- ## **Goals:** 1. Find the matrix \( M_b \) of \( M \) in basis \( b \). 2. Find the vector \( v_b \), the coordinate of \( v \) in basis \( b \). --- ## **Step 1: Find the coordinates of \( v \) in basis \( b \), i.e., \( v_b \).** ### **Express \( v = (1,2,3) \) as a linear combination of \( b_1, b_2, b_3 \):** \[ v = c_1 b_1 + c_2 b_2 + c_3 b_3 \] which gives: \[ (1,2,3) = c_1 (1,1,0) + c_2 (0,1,1) + c_3 (1,0,1) \] This leads to the system: \[ \begin{cases} 1 = c_1 \times 1 + c_2 \times 0 + c_3 \times 1 \\ 2 = c_1 \times 1 + c_2 \times 1 + c_3 \times 0 \\ 3 = c_1 \times 0 + c_2 \times 1 + c_3 \times 1 \end{cases} \] Simplify: \[ \begin{cases} 1 = c_1 + c_3 \quad (1) \\ 2 = c_1 + c_2 \quad (2) \\ 3 = c_2 + c_3 \quad (3) \end{cases} \] --- ## **Step 2: Solve for \( c_1, c_2, c_3 \).** From equation (2): \[ c_2 = 2 - c_1 \] From equation (3): \[ 3 = c_2 + c_3 \Rightarrow c_3 = 3 - c_2 \] Substitute \( c_2 = 2 - c_1 \): \[ c_3 = 3 - (2 - c_1) = 3 - 2 + c_1 = 1 + c_1 \] Now, from equation (1): \[ 1 = c_1 + c_3 = c_1 + (1 + c_1) = 2 c_1 + 1 \] Solve for \( c_1 \): \[ 2 c_1 + 1 = 1 \Rightarrow 2 c_1 = 0 \Rightarrow c_1 = 0 \] Now, find \( c_2 \): \[ c_2 = 2 - c_1 = 2 - 0 = 2 \] And \( c_3 \): \[ c_3 = 1 + c_1 = 1 + 0 = 1 \] ### **Result:** \[ v_b = (c_1, c_2, c_3) = (0, 2, 1) \] --- ## **Step 3: Find the matrix \( M_b \) of \( M \) in basis \( b \).** ### **Method:** - To find \( M_b \), compute \( M \) acting on each basis vector \( b_i \), then express the result in basis \( b \). - The columns of \( M_b \) are the coordinates of \( M b_i \) in basis \( b \). --- ## **Step 4: Compute \( M b_1, M b_2, M b_3 \).** ### **Recall:** \[ M e_j = \text{column } j \text{ of } M \] From the problem: \[ M e = \left[ \begin{bmatrix} 1 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & -1 & 1 \end{bmatrix} \right] e \] Thus, \[ M e_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}, \quad M e_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \quad M e_3 = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} \] --- ## **Step 5: Express \( M b_i \) in standard basis and then in basis \( b \).** ### **1. Compute \( M b_1 \):** \[ b_1 = (1,1,0) \] \[ M b_1 = M (1,1,0) = 1 \times M e_1 + 1 \times M e_2 + 0 \times M e_3 \] \[ M b_1 = 1 \times \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + 1 \times \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + 0 = \begin{bmatrix} 1+1 \\ 2+0 \\ 0-1 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ -1 \end{bmatrix} \] ### **2. Compute \( M b_2 \):** \[ b_2 = (0,1,1) \] \[ M b_2 = 0 \times M e_1 + 1 \times M e_2 + 1 \times M e_3 = 0 + \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1+0 \\ 0+2 \\ -1+1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} \] ### **3. Compute \( M b_3 \):** \[ b_3 = (1,0,1) \] \[ M b_3 = 1 \times M e_1 + 0 \times M e_2 + 1 \times M e_3 = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} + 0 + \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1+0 \\ 2+2 \\ 0+1 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ 1 \end{bmatrix} \] --- ## **Step 6: Express \( M b_i \) in the basis \( b \).** Recall basis vectors: \[ b_1 = (1,1,0), \quad b_2 = (0,1,1), \quad b_3 = (1,0,1) \] Express each \( M b_i \) as a linear combination: \[ M b_i = \alpha_{i1} b_1 + \alpha_{i2} b_2 + \alpha_{i3} b_3 \] **Set up equations for each:** ### **a. For \( M b_1 = (2,2,-1) \):** \[ (2,2,-1) = \alpha_{11}(1,1,0) + \alpha_{12}(0,1,1) + \alpha_{13}(1,0,1) \] which gives: \[ \begin{cases} 2 = \alpha_{11} \times 1 + \alpha_{12} \times 0 + \alpha_{13} \times 1 \Rightarrow 2 = \alpha_{11} + \alpha_{13} \quad (a) \\ 2 = \alpha_{11} \times 1 + \alpha_{12} \times 1 + \alpha_{13} \times 0 \Rightarrow 2 = \alpha_{11} + \alpha_{12} \quad (b) \\ -1 = \alpha_{11} \times 0 + \alpha_{12} \times 1 + \alpha_{13} \times 1 \Rightarrow -1 = \alpha_{12} + \alpha_{13} \quad (c) \end{cases} \] Solve: From (a): \(\alpha_{11} = 2 - \alpha_{13}\). From (b): \(2 = (2 - \alpha_{13}) + \alpha_{12}\) \[ \Rightarrow \alpha_{12} = 2 - (2 - \alpha_{13}) = \alpha_{13} \] From (c): \(-1 = \alpha_{12} + \alpha_{13} = \alpha_{13} + \alpha_{13} = 2 \alpha_{13}\) \[ \Rightarrow 2 \alpha_{13} = -1 \Rightarrow \alpha_{13} = -\frac{1}{2} \] Now, \(\alpha_{12} = -\frac{1}{2}\). And \(\alpha_{11} = 2 - (-\frac{1}{2}) = 2 + \frac{1}{2} = \frac{5}{2}\). ### **b. For \( M b_2 = (1,2,0) \):** \[ (1,2,0) = \beta_{21} b_1 + \beta_{22} b_2 + \beta_{23} b_3 \] Set up equations: \[ \begin{cases} 1 = \beta_{21} + 0 + \beta_{23} \quad (d) \\ 2 = \beta_{21} + \beta_{22} + 0 \quad (e) \\ 0 = 0 + \beta_{22} + \beta_{23} \quad (f) \end{cases} \] From (d): \(\beta_{23} = 1 - \beta_{21}\). From (e): \(\beta_{22} = 2 - \beta_{21}\). From (f): \[ 0 = \beta_{22} + \beta_{23} = (2 - \beta_{21}) + (1 - \beta_{21}) = 3 - 2 \beta_{21} \] Solve: \[ 3 - 2 \beta_{21} = 0 \Rightarrow 2 \beta_{21} = 3 \Rightarrow \beta_{21} = \frac{3}{2} \] Then, \[ \beta_{22} = 2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{1}{2} \] And \[ \beta_{23} = 1 - \frac{3}{2} = -\frac{1}{2} \] ### **c. For \( M b_3 = (1,4,1) \):** \[ (1,4,1) = \gamma_{31} b_1 + \gamma_{32} b_2 + \gamma_{33} b_3 \] Set up equations: \[ \begin{cases} 1 = \gamma_{31} + 0 + \gamma_{33} \quad (g) \\ 4 = \gamma_{31} + \gamma_{32} + 0 \quad (h) \\ 1 = 0 + \gamma_{32} + \gamma_{33} \quad (i) \end{cases} \] From (g): \(\gamma_{33} = 1 - \gamma_{31}\). From (h): \(\gamma_{32} = 4 - \gamma_{31}\). From (i): \[ 1 = \gamma_{32} + \gamma_{33} = (4 - \gamma_{31}) + (1 - \gamma_{31}) = 5 - 2 \gamma_{31} \] Solve: \[ 5 - 2 \gamma_{31} = 1 \Rightarrow 2 \gamma_{31} = 4 \Rightarrow \gamma_{31} = 2 \] Then, \[ \gamma_{32} = 4 - 2 = 2 \] and \[ \gamma_{33} = 1 - 2 = -1 \] --- ## **Step 7: Assemble \( M_b \).** The matrix \( M_b \) has columns: \[ M_b = \begin{bmatrix} \text{