| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the orthogonal projection of the vector \(\mathbf{v} = 9\mathbf{e}_1\) onto the subspace \(W\) of \(\mathbb{R}^4\) spanned by the vectors \(\mathbf{u}_1 = [2, 2, 1, 0]\) and \(\mathbf{u}_2 = [-2, 2, 0, 1]\), we can follow these steps: 1. **Express \(\mathbf{v}\) as a linear combination of \(\mathbf{u}_1\) and \(\mathbf{u}_2\)**, if possible. However, since \(\mathbf{v}\) is not in the span of \(\mathbf{u}_1\) and \(\mathbf{u}_2\), we will instead find the projection of \(\mathbf{v}\) onto \(W\). 2. **Find an orthonormal basis for \(W\)**. To do this, we can use the Gram-Schmidt process to find an orthonormal basis from \(\mathbf{u}_1\) and \(\mathbf{u}_2\). 3. **Use the orthonormal basis to find the projection of \(\mathbf{v}\) onto \(W\)** using the formula \(\text{proj}_W(\mathbf{v}) = (\mathbf{v} \cdot \mathbf{q}_1)\mathbf{q}_1 + (\mathbf{v} \cdot \mathbf{q}_2)\mathbf{q}_2\), where \(\mathbf{q}_1\) and \(\mathbf{q}_2\) are the orthonormal basis vectors. Let's perform these steps: ### Step 1: Orthonormal Basis 1.1. **Find \(\mathbf{q}_1\)**: Normalize \(\mathbf{u}_1\). \[ \mathbf{q}_1 = \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|} = \frac{[2, 2, 1, 0]}{\sqrt{2^2 + 2^2 + 1^2 + 0^2}} = \frac{[2, 2, 1, 0]}{3} \] 1.2. **Find \(\mathbf{q}_2\)**: Find a vector orthogonal to \(\mathbf{q}_1\) and then normalize it. \[ \mathbf{q}_2 = \frac{\mathbf{u}_2 - \text{proj}_{\mathbf{q}_1}(\mathbf{u}_2)}{\|\mathbf{u}_2 - \text{proj}_{\mathbf{q}_1}(\mathbf{u}_2)\|} \] First, find \(\text{proj}_{\mathbf{q}_1}(\mathbf{u}_2)\): \[ \text{proj}_{\mathbf{q}_1}(\mathbf{u}_2) = \left(\frac{\mathbf{u}_2 \cdot \mathbf{q}_1}{\mathbf{q}_1 \cdot \mathbf{q}_1}\right)\mathbf{q}_1 = \left(\frac{[2, 2, 0, 1] \cdot [2, 2, 1, 0]}{3^2}\right)\mathbf{q}_1 \] \[ = \left(\frac{4 + 4}{9}\right)\mathbf{q}_1 = \frac{8}{9}\mathbf{q}_1 \] Then, find \(\mathbf{q}_2\): \[ \mathbf{q}_2 = \frac{\mathbf{u}_2 - \frac{8}{9}\mathbf{q}_1}{\|\mathbf{u}_2 - \frac{8}{9}\mathbf{q}_1\|} \] ### Step 2: Find the Projection 2.1. **Calculate the projection of \(\mathbf{v}\) onto \(W\)**: \[ \text{proj}_W(\mathbf{v}) = (\mathbf{v} \cdot \mathbf{q}_1)\mathbf{q}_1 + (\mathbf{v} \cdot \mathbf{q}_2)\mathbf{q}_2 \] We know \(\mathbf{v} = [9, 0, 0, 0]\), so: \[ (\mathbf{v} \cdot \mathbf{q}_1) = [9, 0, 0, 0] \cdot \left[\frac{2}{3}, \frac{2}{3}, \frac{1}{3}, 0\right] = 6 \] \[ (\mathbf{v} \cdot \mathbf{q}_2) = [9, 0, 0, 0] \cdot \mathbf{q}_2 = 0 \] Thus, the projection of \(\mathbf{v}\) onto \(W\) is: \[ \text{proj}_W(\mathbf{v}) = 6\mathbf{q}_1 + 0\mathbf{q}_2 = 6\left[\frac{2}{3}, \frac{2}{3}, \frac{1}{3}, 0\right] = [4, 4, 2, 0] \] **Answer**: The orthogonal projection of \(9\mathbf{e}_1\) onto the subspace of \(\mathbb{R}^4\) spanned by \([2, 2, 1, 0]\) and \([-2, 2, 0, 1]\) is \([4, 4, 2, 0]\). |
[8, 0, 2, -2] |
| deepseekr1 | To find the orthogonal projection of \(9\mathbf{e}_1\) onto the subspace of \(\mathbb{R}^4\) spanned by \([2, 2, 1, 0]\) and \([-2, 2, 0, 1]\), we follow these steps: 1. **Form the matrix \(A\) with the spanning vectors as columns:** \[ A = \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \] 2. **Compute \(A^T A\):** \[ A^T A = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \] 3. **Compute \(A^T b\) where \(b = [9, 0, 0, 0]^T\):** \[ A^T b = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 9 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 18 \\ -18 \end{bmatrix} \] 4. **Solve the normal equations \((A^T A)x = A^T b\):** \[ \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 18 \\ -18 \end{bmatrix} \] Multiplying both sides by the inverse of \(A^T A\): \[ x = \begin{bmatrix} \frac{1}{9} & 0 \\ 0 & \frac{1}{9} \end{bmatrix} \begin{bmatrix} 18 \\ -18 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix} \] 5. **Compute the projection vector using the coefficients \(x_1 = 2\) and \(x_2 = -2\):** \[ \text{Projection} = 2 \begin{bmatrix} 2 \\ 2 \\ 1 \\ 0 \end{bmatrix} + (-2) \begin{bmatrix} -2 \\ 2 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ 4 \\ 2 \\ 0 \end{bmatrix} + \begin{bmatrix} 4 \\ -4 \\ 0 \\ -2 \end{bmatrix} = \begin{bmatrix} 8 \\ 0 \\ 2 \\ -2 \end{bmatrix} \] The orthogonal projection of \(9\mathbf{e}_1\) onto the subspace is \(\boxed{\left[8,\ 0,\ 2,\ -2\right]}\). |
[8, 0, 2, -2] |
| gpto1 | To find the orthogonal projection of the vector \( y = 9\mathbf{e}_1 = [9, 0, 0, 0] \) onto the subspace \( S \) of \( \mathbb{R}^4 \) spanned by the vectors \[ \mathbf{u} = [2, 2, 1, 0], \quad \mathbf{v} = [-2, 2, 0, 1], \] we can use the projection formula involving matrices. **Step 1:** Build the matrix \( A \) whose columns are the vectors spanning \( S \): \[ A = [\mathbf{u} \quad \mathbf{v}] = \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \\ \end{bmatrix}. \] **Step 2:** Compute \( A^T A \): \[ A^T A = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \\ \end{bmatrix}. \] **Step 3:** Compute \( A^T y \): \[ A^T y = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 9 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} (2)(9) + (2)(0) + (1)(0) + (0)(0) \\ (-2)(9) + (2)(0) + (0)(0) + (1)(0) \\ \end{bmatrix} = \begin{bmatrix} 18 \\ -18 \\ \end{bmatrix}. \] **Step 4:** Compute \( (A^T A)^{-1} \): \[ (A^T A)^{-1} = \begin{bmatrix} 1/9 & 0 \\ 0 & 1/9 \\ \end{bmatrix}. \] **Step 5:** Solve for \( c \): \[ c = (A^T A)^{-1} A^T y = \begin{bmatrix} 1/9 & 0 \\ 0 & 1/9 \\ \end{bmatrix} \begin{bmatrix} 18 \\ -18 \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \\ \end{bmatrix}. \] **Step 6:** Compute the projection \( \text{proj}_S(y) \): \[ \text{proj}_S(y) = A c = [\mathbf{u} \quad \mathbf{v}] \begin{bmatrix} 2 \\ -2 \\ \end{bmatrix} = 2\mathbf{u} - 2\mathbf{v} = 2[2, 2, 1, 0] - 2[-2, 2, 0, 1]. \] **Step 7:** Perform the vector operations: \[ 2\mathbf{u} = [4, 4, 2, 0], \quad 2\mathbf{v} = [-4, 4, 0, 2]. \] \[ \text{proj}_S(y) = [4, 4, 2, 0] - [-4, 4, 0, 2] = [4 + 4, 4 - 4, 2 - 0, 0 - 2] = [8, 0, 2, -2]. \] **Answer:** The orthogonal projection of 9e₁ onto the subspace is [8, 0, 2, –2] |
[8, 0, 2, -2] |
| GPT-4o | To find the orthogonal projection of the vector \( \mathbf{v} = 9\mathbf{e}_1 = [9, 0, 0, 0] \) onto the subspace \( W \) of \( \mathbb{R}^4 \) spanned by the vectors \( \mathbf{u}_1 = [2, 2, 1, 0] \) and \( \mathbf{u}_2 = [-2, 2, 0, 1] \), we can follow these steps: 1. **Form the matrix \( A \) with \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) as columns:** \[ A = \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \] 2. **Compute the Gram matrix \( G \) which is \( A^T A \):** \[ G = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \] Calculating each element of \( G \): \[ G_{11} = (2)(2) + (2)(2) + (1)(1) + (0)(0) = 4 + 4 + 1 = 9 \] \[ G_{12} = (2)(-2) + (2)(2) + (1)(0) + (0)(1) = -4 + 4 + 0 + 0 = 0 \] \[ G_{21} = (-2)(2) + (2)(2) + (0)(1) + (1)(0) = -4 + 4 + 0 + 0 = 0 \] \[ G_{22} = (-2)(-2) + (2)(2) + (0)(0) + (1)(1) = 4 + 4 + 0 + 1 = 9 \] Therefore, \[ G = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \] 3. **Compute \( A^T \mathbf{v} \):** \[ A^T \mathbf{v} = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 9 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} (2)(9) + (2)(0) + (1)(0) + (0)(0) \\ (-2)(9) + (2)(0) + (0)(0) + (1)(0) \end{bmatrix} = \begin{bmatrix} 18 \\ -18 \end{bmatrix} \] 4. **Solve for \( \mathbf{c} \) in \( G \mathbf{c} = A^T \mathbf{v} \):** \[ \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} 18 \\ -18 \end{bmatrix} \] This gives us: \[ 9c_1 = 18 \implies c_1 = 2 \] \[ 9c_2 = -18 \implies c_2 = -2 \] 5 |
[8, 0, 2, -2] |
| llama3.1-70B-Instruct | To find the orthogonal projection of 9e_1 onto the subspace of R^4 spanned by [2, 2, 1, 0] and [-2, 2, 0, 1], we can follow these steps: 1. Find the basis for the subspace: The subspace is spanned by two vectors: [2, 2, 1, 0] and [-2, 2, 0, 1]. We can check if these vectors are linearly independent by computing their dot product: [2, 2, 1, 0] · [-2, 2, 0, 1] = -4 + 4 + 0 + 0 = 0 Since the dot product is zero, the vectors are orthogonal, and we can conclude that they form a basis for the subspace. 2. Find the orthogonal projection of 9e_1 onto the subspace: Let's denote the basis vectors as v_1 = [2, 2, 1, 0] and v_2 = [-2, 2, 0, 1]. We can compute the orthogonal projection of 9e_1 onto the subspace using the formula: proj_v(9e_1) = (9e_1 · v_1) / ||v_1||^2 \* v_1 + (9e_1 · v_2) / ||v_2||^2 \* v_2 First, we compute the dot products: 9e_1 · v_1 = 9 \* [1, 0, 0, 0] · [2, 2, 1, 0] = 18 9e_1 · v_2 = 9 \* [1, 0, 0, 0] · [-2, 2, 0, 1] = -18 Next, we compute the norms: ||v_1||^2 = [2, 2, 1, 0] · [2, 2, 1, 0] = 4 + 4 + 1 + 0 = 9 ||v_2||^2 = [-2, 2, 0, 1] · [-2, 2, 0, 1] = 4 + 4 + 0 + 1 = 9 Now, we can plug in the values: proj_v(9e_1) = (18 / 9) \* [2, 2, 1, 0] + (-18 / 9) \* [-2, 2, 0, 1] = 2 \* [2, 2, 1, 0] - 2 \* [-2, 2, 0, 1] = [4, 4, 2, 0] - [-4, 4, 0, 2] = [8, 0, 2, -2] Therefore, the orthogonal projection of 9e_1 onto the subspace of R^4 spanned by [2, 2, 1, 0] and [-2, 2, 0, 1] is [8, 0, 2, -2]. |
[8, 0, 2, -2] |
| llama3.1-405B-Instruct | To find the orthogonal projection of 9e_1 onto the subspace of R^4 spanned by [2, 2, 1, 0] and [-2, 2, 0, 1], we can use the formula for the orthogonal projection of a vector onto a subspace. Let's denote the subspace as W and the vector 9e_1 as v. We need to find the orthogonal projection of v onto W, denoted as proj_W(v). First, we need to find an orthonormal basis for W. We can do this by applying the Gram-Schmidt process to the vectors [2, 2, 1, 0] and [-2, 2, 0, 1]. Let's denote the orthonormal basis vectors as u_1 and u_2. We can calculate them as follows: u_1 = [2, 2, 1, 0] / ||[2, 2, 1, 0]|| = [2, 2, 1, 0] / √(2^2 + 2^2 + 1^2 + 0^2) = [2, 2, 1, 0] / √9 = [2/3, 2/3, 1/3, 0] u_2 = [-2, 2, 0, 1] - ([-2, 2, 0, 1] \* u_1) \* u_1 = [-2, 2, 0, 1] - ([-2, 2, 0, 1] \* [2/3, 2/3, 1/3, 0]) \* [2/3, 2/3, 1/3, 0] = [-2, 2, 0, 1] - [-4/3, 4/3, 0, 0] \* [2/3, 2/3, 1/3, 0] = [-2, 2, 0, 1] - [-8/9, 8/9, 0, 0] = [-2/9, 2/9, 0, 1] Now we have an orthonormal basis for W: u_1 = [2/3, 2/3, 1/3, 0] and u_2 = [-2/9, 2/9, 0, 1]. Next, we can calculate the orthogonal projection of v onto W using the formula: proj_W(v) = (v \* u_1) \* u_1 + (v \* u_2) \* u_2 where v = 9e_1 = [9, 0, 0, 0]. First, we calculate the dot products: v \* u_1 = [9, 0, 0, 0] \* [2/3, 2/3, 1/3, 0] = 18/3 = 6 v \* u_2 = [9, 0, 0, 0] \* [-2/9, 2/9, 0, 1] = -18/9 = -2 Now we can calculate the orthogonal projection: proj_W(v) = (6) \* [2/3, 2/3, 1/3, 0] + (-2) \* [-2/9, 2/9, 0, 1] = [4, 4, 2, 0] + [4/9, -4/9, 0, -2] = [4 + 4/9, 4 - 4/9, 2, -2] = [40/9, 32/9, 2, -2] Therefore, the orthogonal projection of 9e_1 onto the subspace of R^4 spanned by [2, 2, 1, 0] and [-2, 2, 0, 1] is [40/9, 32/9, 2, -2]. |
[8, 0, 2, -2] |
| DeepSeek-Coder-V2 | To find the orthogonal projection of \(9e_1\) onto the subspace of \(\mathbb{R}^4\) spanned by the vectors \([2, 2, 1, 0]\) and \([-2, 2, 0, 1]\), we can follow these steps: 1. **Identify the given vectors and the vector to be projected:** - The subspace is spanned by the vectors \(v_1 = [2, 2, 1, 0]\) and \(v_2 = [-2, 2, 0, 1]\). - The vector to be projected is \(9e_1 = [9, 0, 0, 0]\). 2. **Form the matrix \(A\) whose columns are the spanning vectors:** \[ A = \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \] 3. **Compute the projection of \(9e_1\) onto the subspace:** The orthogonal projection formula is given by: \[ \text{proj}_{\text{subspace}} (9e_1) = A(A^T A)^{-1} A^T (9e_1) \] 4. **Calculate \(A^T A\):** \[ A^T A = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 \cdot 2 + 2 \cdot 2 + 1 \cdot 1 + 0 \cdot 0 & 2 \cdot -2 + 2 \cdot 2 + 1 \cdot 0 + 0 \cdot 1 \\ -2 \cdot 2 + 2 \cdot 2 + 0 \cdot 1 + 1 \cdot 0 & -2 \cdot -2 + 2 \cdot 2 + 0 \cdot 0 + 1 \cdot 1 \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix} \] 5. **Compute \((A^T A)^{-1}\):** \[ (A^T A)^{-1} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}^{-1} = \begin{bmatrix} \frac{1}{9} & 0 \\ 0 & \frac{1}{9} \end{bmatrix} \] 6. **Calculate \(A(A^T A)^{-1} A^T\):** \[ A(A^T A)^{-1} A^T = \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{9} & 0 \\ 0 & \frac{1}{9} \end{bmatrix} \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{9} & -\frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} \\ \frac{1}{9} & 0 \\ 0 & \frac{1}{9} \end{bmatrix} \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \end{bmatrix} \] 7. **Multiply the matrices:** \[ = \begin{bmatrix} \frac{4}{9} - \frac{4}{9} & \frac{4}{9} + \frac{4}{9} & \frac{2}{9} & -\frac{2}{9} \\ \frac{4}{9} + \frac{4}{9} & \frac{4}{9} + \frac{4}{9} & \frac{2}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} & 0 \\ -\frac{2}{9} & \frac{2}{9} & 0 & \frac{1}{9} \end{bmatrix} = \begin{bmatrix} 0 & \frac{8}{9} & \frac{2}{9} & -\frac{2}{9} \\ \frac{8}{9} & 0 & \frac{2}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} & 0 \\ -\frac{2}{9} & \frac{2}{9} & 0 & \frac{1}{9} \end{bmatrix} \] 8. **Compute the projection:** \[ \text{proj}_{\text{subspace}} (9e_1) = \begin{bmatrix} 0 & \frac{8}{9} & \frac{2}{9} & -\frac{2}{9} \\ \frac{8}{9} & 0 & \frac{2}{9} & \frac{2}{9} \\ \frac{2}{9} & \frac{2}{9} & \frac{1}{9} & 0 \\ -\frac{2}{9} & \frac{2}{9} & 0 & \frac{1}{9} \end{bmatrix} \begin{bmatrix} 9 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{8}{9} \cdot 9 \\ \frac{2}{9} \cdot 9 \\ -\frac{2}{9} \cdot 9 \end{bmatrix} = \begin{bmatrix} 0 \\ 8 \\ 2 \\ -2 \end{bmatrix} \] Thus, the orthogonal projection of \(9e_1\) onto the subspace spanned by \([2, 2, 1, 0]\) and \([-2, 2, 0, 1]\) is: \[ \boxed{[0, 8, 2, -2]} \] |
[8, 0, 2, -2] |
| Qwen2-72B-Instruct | To find the orthogonal projection of a vector \(\mathbf{v} = 9\mathbf{e}_1\) onto a subspace \(W\) of \(\mathbb{R}^4\) spanned by vectors \(\mathbf{u}_1 = [2, 2, 1, 0]\) and \(\mathbf{u}_2 = [-2, 2, 0, 1]\), we can use the following steps: 1. **Express \(\mathbf{v}\) as a linear combination of \(\mathbf{u}_1\) and \(\mathbf{u}_2\)**, if possible. However, since \(\mathbf{v}\) is not in the span of \(\mathbf{u}_1\) and \(\mathbf{u}_2\), we will instead find the projection of \(\mathbf{v}\) onto \(W\). 2. **Find an orthonormal basis for \(W\)**. To do this, we can use the Gram-Schmidt process to find an orthonormal basis from \(\mathbf{u}_1\) and \(\mathbf{u}_2\). 3. **Use the orthonormal basis to find the projection of \(\mathbf{v}\) onto \(W\)** using the formula \(\text{proj}_W(\mathbf{v}) = (\mathbf{v} \cdot \mathbf{q}_1)\mathbf{q}_1 + (\mathbf{v} \cdot \mathbf{q}_2)\mathbf{q}_2\), where \(\mathbf{q}_1\) and \(\mathbf{q}_2\) are the orthonormal basis vectors. Let's perform these steps: ### Step 1: Orthonormal Basis 1.1. **Find \(\mathbf{q}_1\)**: Normalize \(\mathbf{u}_1\). \[ \mathbf{q}_1 = \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|} = \frac{[2, 2, 1, 0]}{\sqrt{2^2 + 2^2 + 1^2 + 0^2}} = \frac{[2, 2, 1, 0]}{3} \] 1.2. **Find \(\mathbf{q}_2\)**: Find a vector orthogonal to \(\mathbf{q}_1\) and then normalize it. \[ \mathbf{q}_2 = \frac{\mathbf{u}_2 - \text{proj}_{\mathbf{q}_1}(\mathbf{u}_2)}{\|\mathbf{u}_2 - \text{proj}_{\mathbf{q}_1}(\mathbf{u}_2)\|} \] First, find \(\text{proj}_{\mathbf{q}_1}(\mathbf{u}_2)\): \[ \text{proj}_{\mathbf{q}_1}(\mathbf{u}_2) = \left(\frac{\mathbf{u}_2 \cdot \mathbf{q}_1}{\mathbf{q}_1 \cdot \mathbf{q}_1}\right)\mathbf{q}_1 = \left(\frac{[2, 2, 0, 1] \cdot [2, 2, 1, 0]}{3^2}\right)\mathbf{q}_1 \] \[ = \left(\frac{4 + 4}{9}\right)\mathbf{q}_1 = \frac{8}{9}\mathbf{q}_1 \] Then, find \(\mathbf{q}_2\): \[ \mathbf{q}_2 = \frac{\mathbf{u}_2 - \frac{8}{9}\mathbf{q}_1}{\|\mathbf{u}_2 - \frac{8}{9}\mathbf{q}_1\|} \] ### Step 2: Find the Projection 2.1. **Calculate the projection of \(\mathbf{v}\) onto \(W\)**: \[ \text{proj}_W(\mathbf{v}) = (\mathbf{v} \cdot \mathbf{q}_1)\mathbf{q}_1 + (\mathbf{v} \cdot \mathbf{q}_2)\mathbf{q}_2 \] We know \(\mathbf{v} = [9, 0, 0, 0]\), so: \[ (\mathbf{v} \cdot \mathbf{q}_1) = [9, 0, 0, 0] \cdot \left[\frac{2}{3}, \frac{2}{3}, \frac{1}{3}, 0\right] = 6 \] \[ (\mathbf{v} \cdot \mathbf{q}_2) = [9, 0, 0, 0] \cdot \mathbf{q}_2 = 0 \] Thus, the projection of \(\mathbf{v}\) onto \(W\) is: \[ \text{proj}_W(\mathbf{v}) = 6\mathbf{q}_1 + 0\mathbf{q}_2 = 6\left[\frac{2}{3}, \frac{2}{3}, \frac{1}{3}, 0\right] = [4, 4, 2, 0] \] **Answer**: The orthogonal projection of \(9\mathbf{e}_1\) onto the subspace of \(\mathbb{R}^4\) spanned by \([2, 2, 1, 0]\) and \([-2, 2, 0, 1]\) is \([4, 4, 2, 0]\). |
[8, 0, 2, -2] |
| baichuan4 | To find the orthogonal projection of a vector onto a subspace, we can use the formula: Projection(v) = (v ⋅ u1) / (u1 ⋅ u1) * u1 + (v ⋅ u2) / (u2 ⋅ u2) * u2 where v is the vector we want to project, u1 and u2 are the basis vectors of the subspace, and ⋅ denotes the dot product. In this case, v = 9e_1 = [9, 0, 0, 0], u1 = [2, 2, 1, 0], and u2 = [-2, 2, 0, 1]. First, we need to calculate the dot products: v ⋅ u1 = 9*2 + 0*2 + 0*1 + 0*0 = 18 v ⋅ u2 = 9*(-2) + 0*2 + 0*0 + 0*1 = -18 u1 ⋅ u1 = 2*2 + 2*2 + 1*1 + 0*0 = 9 u2 ⋅ u2 = (-2)*(-2) + 2*2 + 0*0 + 1*1 = 9 Now, we can plug these values into the formula: Projection(9e_1) = (18 / 9) * [2, 2, 1, 0] + (-18 / 9) * [-2, 2, 0, 1] = 2 * [2, 2, 1, 0] - 2 * [-2, 2, 0, 1] = [4, 4, 2, 0] - [-4, -4, 0, -2] = [4 + 4, 4 - 4, 2 - 0, 0 - (-2)] = [8, 0, 2, 2] So, the orthogonal projection of 9e_1 onto the subspace spanned by [2, 2, 1, 0] and [-2, 2, 0, 1] is the vector [8, 0, 2, 2]. |
[8, 0, 2, -2] |
| doubao | Let's call the vectors spanning the subspace as $v_1 = [2, 2, 1, 0]$ and $v_2 = [-2, 2, 0, 1]$. First find the orthonormal basis for the subspace. Normalize $v_1$ to get $u_1 = \frac{v_1}{\|v_1\|} = \frac{1}{3}[2, 2, 1, 0]$. Find the projection of $v_2$ onto $u_1$: $proj_{u_1}(v_2) = (v_2 \cdot u_1)u_1 = (\frac{-4+4+0+0}{3})u_1 = 0$. So $v_2$ is orthogonal to $u_1$. Normalize $v_2$ to get $u_2 = \frac{v_2}{\|v_2\|} = \frac{1}{3}[-2, 2, 0, 1]$. Now the orthogonal projection of $9e_1 = [9, 0, 0, 0]$ onto the subspace is given by: $proj(9e_1) = (9e_1 \cdot u_1)u_1 + (9e_1 \cdot u_2)u_2 = (9 \cdot \frac{2}{3})\frac{1}{3}[2, 2, 1, 0] + (9 \cdot 0)\frac{1}{3}[-2, 2, 0, 1] = 2[2, 2, 1, 0] = [4, 4, 2, 0]$. |
[8, 0, 2, -2] |
| minimax | To find the orthogonal projection of a vector \(\mathbf{v}\) onto a subspace \(W\) spanned by vectors \(\mathbf{u}_1\) and \(\mathbf{u}_2\), we can use the formula for the projection of \(\mathbf{v}\) onto \(W\): \[ \text{proj}_W(\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 + \frac{\mathbf{v} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2 \] where \(\cdot\) denotes the dot product of two vectors. Given vectors \(\mathbf{v} = [9, 0, 0, 0]\), \(\mathbf{u}_1 = [2, 2, 1, 0]\), and \(\mathbf{u}_2 = [-2, 2, 0, 1]\), we can calculate the dot products: \[ \mathbf{v} \cdot \mathbf{u}_1 = 9 \cdot 2 + 0 \cdot 2 + 0 \cdot 1 + 0 \cdot 0 = 18 \] \[ \mathbf{u}_1 \cdot \mathbf{u}_1 = 2^2 + 2^2 + 1^2 + 0^2 = 4 + 4 + 1 = 9 \] \[ \mathbf{v} \cdot \mathbf{u}_2 = 9 \cdot (-2) + 0 \cdot 2 + 0 \cdot 0 + 0 \cdot 1 = -18 \] \[ \mathbf{u}_2 \cdot \mathbf{u}_2 = (-2)^2 + 2^2 + 0^2 + 1^2 = 4 + 4 + 0 + 1 = 9 \] Now we can compute the orthogonal projection: \[ \text{proj}_W(\mathbf{v}) = \frac{18}{9} \mathbf{u}_1 + \frac{-18}{9} \mathbf{u}_2 \] \[ \text{proj}_W(\mathbf{v}) = 2 \mathbf{u}_1 - 2 \mathbf{u}_2 \] \[ \text{proj}_W(\mathbf{v}) = 2[2, 2, 1, 0] - 2[-2, 2, 0, 1] \] \[ \text{proj}_W(\mathbf{v}) = [4, 4, 2, 0] - [-4, 4, 0, 2] \] \[ \text{proj}_W(\mathbf{v}) = [4 + 4, 4 - 4, 2 - 0, 0 - 2] \] \[ \text{proj}_W(\mathbf{v}) = [8, 0, 2, -2] \] Therefore, the orthogonal projection of \(9\mathbf{e}_1\) onto the subspace spanned by \([2, 2, 1, 0]\) and \([-2, 2, 0, 1]\) is \([8, 0, 2, -2]\). |
[8, 0, 2, -2] |
| kimi | To find the orthogonal projection of a vector \( \mathbf{v} \) onto a subspace spanned by vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), we can use the following formula: \[ \text{proj}_{\mathbf{u}_1, \mathbf{u}_2} \mathbf{v} = \left( \frac{\mathbf{v} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \right) \mathbf{u}_1 + \left( \frac{\mathbf{v} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \right) \mathbf{u}_2 \] Given: \( \mathbf{v} = 9\mathbf{e}_1 = \begin{bmatrix} 9 \\ 0 \\ 0 \\ 0 \end{bmatrix} \) \( \mathbf{u}_1 = \begin{bmatrix} 2 \\ 2 \\ 1 \\ 0 \end{bmatrix} \) \( \mathbf{u}_2 = \begin{bmatrix} -2 \\ 2 \\ 0 \\ 1 \end{bmatrix} \) First, we calculate the dot products \( \mathbf{v} \cdot \mathbf{u}_1 \), \( \mathbf{u}_1 \cdot \mathbf{u}_1 \), \( \mathbf{v} \cdot \mathbf{u}_2 \), and \( \mathbf{u}_2 \cdot \mathbf{u}_2 \). \[ \mathbf{v} \cdot \mathbf{u}_1 = 9 \cdot 2 = 18 \] \[ \mathbf{u}_1 \cdot \mathbf{u}_1 = 2^2 + 2^2 + 1^2 + 0^2 = 4 + 4 + 1 + 0 = 9 \] \[ \mathbf{v} \cdot \mathbf{u}_2 = 9 \cdot (-2) = -18 \] \[ \mathbf{u}_2 \cdot \mathbf{u}_2 = (-2)^2 + 2^2 + 0^2 + 1^2 = 4 + 4 + 0 + 1 = 9 \] Now, we can find the coefficients for the projection: \[ \frac{\mathbf{v} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} = \frac{18}{9} = 2 \] \[ \frac{\mathbf{v} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} = \frac{-18}{9} = -2 \] Finally, we multiply these coefficients by the respective vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) and add the results to find the projection: \[ \text{proj}_{\mathbf{u}_1, \mathbf{u}_2} \mathbf{v} = 2 \mathbf{u}_1 - 2 \mathbf{u}_2 = 2 \begin{bmatrix} 2 \\ 2 \\ 1 \\ 0 \end{bmatrix} - 2 \begin{bmatrix} -2 \\ 2 \\ 0 \\ 1 \end{bmatrix} \] \[ = \begin{bmatrix} 4 \\ 4 \\ 2 \\ 0 \end{bmatrix} - \begin{bmatrix} -4 \\ 4 \\ 0 \\ 2 \end{bmatrix} \] \[ = \begin{bmatrix} 4 + 4 \\ 4 - 4 \\ 2 - 0 \\ 0 - 2 \end{bmatrix} \] \[ = \begin{bmatrix} 8 \\ 0 \\ 2 \\ -2 \end{bmatrix} \] So, the orthogonal projection of \( 9\mathbf{e}_1 \) onto the subspace spanned by \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \) is \( \begin{bmatrix} 8 \\ 0 \\ 2 \\ -2 \end{bmatrix} \). |
[8, 0, 2, -2] |
| wenxinyiyan | 首先，我们需要找到向量9e_1在R^4中的表示。由于e_1是标准基向量，它在R^4中的表示为[9, 0, 0, 0]。 接下来，我们需要找到由向量[2, 2, 1, 0]和[-2, 2, 0, 1]张成的子空间的正交基。为了做到这一点，我们可以使用Gram-Schmidt正交化过程。 首先，将向量[2, 2, 1, 0]单位化，得到向量u_1。计算过程如下： u_1 = [2, 2, 1, 0] / sqrt(2^2 + 2^2 + 1^2 + 0^2) = [2, 2, 1, 0] / sqrt(9) = [2/3, 2/3, 1/3, 0] 接下来，我们需要找到与u_1正交的向量u_2。我们可以使用Gram-Schmidt正交化过程，从向量[-2, 2, 0, 1]中减去它在u_1上的投影。计算过程如下： proj_u1_v = ([-2, 2, 0, 1] · u_1) * u_1 = ([-2, 2, 0, 1] · [2/3, 2/3, 1/3, 0]) * [2/3, 2/3, 1/3, 0] = (-4/3 + 4/3 + 0 + 0) * [2/3, 2/3, 1/3, 0] = [0, 0, 0, 0] u_2 = [-2, 2, 0, 1] - proj_u1_v = [-2, 2, 0, 1] - [0, 0, 0, 0] = [-2, 2, 0, 1] 然后，我们将u_2单位化，得到： u_2 = [-2, 2, 0, 1] / sqrt((-2)^2 + 2^2 + 0^2 + 1^2) = [-2, 2, 0, 1] / sqrt(9) = [-2/3, 2/3, 0, 1/3] 现在，我们有了正交基{u_1, u_2}，我们可以找到向量9e_1在这个子空间上的正交投影。要做到这一点，我们只需要计算9e_1在u_1和u_2上的投影，并将它们相加。计算过程如下： proj_u1_9e1 = (9e_1 · u_1) * u_1 = ([9, 0, 0, 0] · [2/3, 2/3, 1/3, 0]) * [2/3, 2/3, 1/3, 0] = (6/3 + 0 + 0 + 0) * [2/3, 2/3, 1/3, 0] = [4/3, 4/3, 2/3, 0] proj_u2_9e1 = (9e_1 · u_2) * u_2 = ([9, 0, 0, 0] · [-2/3, 2/3, 0, 1/3]) * [-2/3, 2/3, 0, 1/3] = (-6/3 + 0 + 0 + 0) * [-2/3, 2/3, 0, 1/3] = [4/3, -4/3, 0, -2/3] 正交投影 = proj_u1_9e1 + proj_u2_9e1 = [4/3, 4/3, 2/3, 0] + [4/3, -4/3, 0, -2/3] = [8/3, 0, 2/3, -2/3] 所以，9e_1在由向量[2, 2, 1, 0]和[-2, 2, 0, 1]张成的子空间上的正交投影是[8/3, |
[8, 0, 2, -2] |
| zhipu | To find the orthogonal projection of a vector onto a subspace, we can use the formula for the projection of a vector \( \mathbf{a} \) onto a subspace spanned by orthogonal vectors \( \mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_k \): \[ \text{proj}_{\text{subspace}} \mathbf{a} = \sum_{i=1}^{k} \frac{\mathbf{a} \cdot \mathbf{u}_i}{\mathbf{u}_i \cdot \mathbf{u}_i} \mathbf{u}_i \] In this case, we are projecting the vector \( 9\mathbf{e}_1 \) (which is \( [9, 0, 0, 0] \) in \( \mathbb{R}^4 \)) onto the subspace spanned by the vectors \( \mathbf{v}_1 = [2, 2, 1, 0] \) and \( \mathbf{v}_2 = [-2, 2, 0, 1] \). First, we need to ensure that \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal. If they are not, we need to orthogonalize them using the Gram-Schmidt process. However, we can check their orthogonality by computing their dot product: \[ \mathbf{v}_1 \cdot \mathbf{v}_2 = 2 \cdot (-2) + 2 \cdot 2 + 1 \cdot 0 + 0 \cdot 1 = -4 + 4 + 0 + 0 = 0 \] Since their dot product is zero, \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal. Now we can proceed with the projection: 1. Compute the projection of \( 9\mathbf{e}_1 \) onto \( \mathbf{v}_1 \): \[ \text{proj}_{\mathbf{v}_1} (9\mathbf{e}_1) = \frac{(9\mathbf{e}_1) \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1 \] \[ (9\mathbf{e}_1) \cdot \mathbf{v}_1 = 9 \cdot 2 + 0 \cdot 2 + 0 \cdot 1 + 0 \cdot 0 = 18 \] \[ \mathbf{v}_1 \cdot \mathbf{v}_1 = 2^2 + 2^2 + 1^2 + 0^2 = 4 + 4 + 1 = 9 \] \[ \text{proj}_{\mathbf{v}_1} (9\mathbf{e}_1) = \frac{18}{9} \mathbf{v}_1 = 2 \mathbf{v}_1 = 2 [2, 2, 1, 0] = [4, 4, 2, 0] \] 2. Compute the projection of \( 9\mathbf{e}_1 \) onto \( \mathbf{v}_2 \): \[ \text{proj}_{\mathbf{v}_2} (9\mathbf{e}_1) = \frac{(9\mathbf{e}_1) \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \mathbf{v}_2 \] \[ (9\mathbf{e}_1) \cdot \mathbf{v}_2 = 9 \cdot (-2) + 0 \cdot 2 + 0 \cdot 0 + 0 \cdot 1 = -18 \] \[ \mathbf{v}_2 \cdot \mathbf{v}_2 = (-2)^2 + 2^2 + 0^2 + 1^2 = 4 + 4 + 0 + 1 = 9 \] \[ \text{proj}_{\mathbf{v}_2} (9\mathbf{e}_1) = \frac{-18}{9} \mathbf{v}_2 = -2 \mathbf{v}_2 = -2 [-2, 2, 0, 1] = [4, -4, 0, -2] \] 3. Add the two projections to get the total projection onto the subspace: \[ \text{proj}_{\text{subspace}} (9\mathbf{e}_1) = \text{proj}_{\mathbf{v}_1} (9\mathbf{e}_1) + \text{proj}_{\mathbf{v}_2} (9\mathbf{e}_1) \] \[ \text{proj}_{\text{subspace}} (9\mathbf{e}_1) = [4, 4, 2, 0] + [4, -4, 0, -2] = [8, 0, 2, -2] \] Therefore, the orthogonal projection of \( 9\mathbf{e}_1 \) onto the subspace of \( \mathbb{R}^4 \) spanned by \( [2, 2, 1, 0] \) and \( [-2, 2, 0, 1] \) is \( [8, 0, 2, -2] \). |
[8, 0, 2, -2] |
| gpto1 | To find the orthogonal projection of the vector \( y = 9\mathbf{e}_1 = [9, 0, 0, 0] \) onto the subspace \( S \) of \( \mathbb{R}^4 \) spanned by the vectors \[ \mathbf{u} = [2, 2, 1, 0], \quad \mathbf{v} = [-2, 2, 0, 1], \] we can use the projection formula involving matrices. **Step 1:** Build the matrix \( A \) whose columns are the vectors spanning \( S \): \[ A = [\mathbf{u} \quad \mathbf{v}] = \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \\ \end{bmatrix}. \] **Step 2:** Compute \( A^T A \): \[ A^T A = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \\ 1 & 0 \\ 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} 9 & 0 \\ 0 & 9 \\ \end{bmatrix}. \] **Step 3:** Compute \( A^T y \): \[ A^T y = \begin{bmatrix} 2 & 2 & 1 & 0 \\ -2 & 2 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 9 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} = \begin{bmatrix} (2)(9) + (2)(0) + (1)(0) + (0)(0) \\ (-2)(9) + (2)(0) + (0)(0) + (1)(0) \\ \end{bmatrix} = \begin{bmatrix} 18 \\ -18 \\ \end{bmatrix}. \] **Step 4:** Compute \( (A^T A)^{-1} \): \[ (A^T A)^{-1} = \begin{bmatrix} 1/9 & 0 \\ 0 & 1/9 \\ \end{bmatrix}. \] **Step 5:** Solve for \( c \): \[ c = (A^T A)^{-1} A^T y = \begin{bmatrix} 1/9 & 0 \\ 0 & 1/9 \\ \end{bmatrix} \begin{bmatrix} 18 \\ -18 \\ \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \\ \end{bmatrix}. \] **Step 6:** Compute the projection \( \text{proj}_S(y) \): \[ \text{proj}_S(y) = A c = [\mathbf{u} \quad \mathbf{v}] \begin{bmatrix} 2 \\ -2 \\ \end{bmatrix} = 2\mathbf{u} - 2\mathbf{v} = 2[2, 2, 1, 0] - 2[-2, 2, 0, 1]. \] **Step 7:** Perform the vector operations: \[ 2\mathbf{u} = [4, 4, 2, 0], \quad 2\mathbf{v} = [-4, 4, 0, 2]. \] \[ \text{proj}_S(y) = [4, 4, 2, 0] - [-4, 4, 0, 2] = [4 + 4, 4 - 4, 2 - 0, 0 - 2] = [8, 0, 2, -2]. \] **Answer:** The orthogonal projection of 9e₁ onto the subspace is [8, 0, 2, –2] |
[8, 0, 2, -2] |
| deepseekv3 | To find the orthogonal projection of the vector \( 9\mathbf{e}_1 = [9, 0, 0, 0] \) onto the subspace of \( \mathbb{R}^4 \) spanned by the vectors \( \mathbf{v}_1 = [2, 2, 1, 0] \) and \( \mathbf{v}_2 = [-2, 2, 0, 1] \), we can use the projection formula for a subspace spanned by an orthogonal basis. ### Step 1: Check if \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal First, we verify whether \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal by computing their dot product: \[ \mathbf{v}_1 \cdot \mathbf{v}_2 = (2)(-2) + (2)(2) + (1)(0) + (0)(1) = -4 + 4 + 0 + 0 = 0. \] Since the dot product is zero, \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are orthogonal, and we can use them directly as an orthogonal basis for the subspace. ### Step 2: Compute the projection The orthogonal projection of \( 9\mathbf{e}_1 \) onto the subspace is given by: \[ \text{proj}_{\text{span}(\mathbf{v}_1, \mathbf{v}_2)} (9\mathbf{e}_1) = \frac{9\mathbf{e}_1 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1 + \frac{9\mathbf{e}_1 \cdot \mathbf{v}_2}{\mathbf{v}_2 \cdot \mathbf{v}_2} \mathbf{v}_2. \] #### Compute \( 9\mathbf{e}_1 \cdot \mathbf{v}_1 \): \[ 9\mathbf{e}_1 \cdot \mathbf{v}_1 = (9)(2) + (0)(2) + (0)(1) + (0)(0) = 18. \] #### Compute \( \mathbf{v}_1 \cdot \mathbf{v}_1 \): \[ \mathbf{v}_1 \cdot \mathbf{v}_1 = (2)^2 + (2)^2 + (1)^2 + (0)^2 = 4 + 4 + 1 + 0 = 9. \] #### Compute \( 9\mathbf{e}_1 \cdot \mathbf{v}_2 \): \[ 9\mathbf{e}_1 \cdot \mathbf{v}_2 = (9)(-2) + (0)(2) + (0)(0) + (0)(1) = -18. \] #### Compute \( \mathbf{v}_2 \cdot \mathbf{v}_2 \): \[ \mathbf{v}_2 \cdot \mathbf{v}_2 = (-2)^2 + (2)^2 + (0)^2 + (1)^2 = 4 + 4 + 0 + 1 = 9. \] #### Substitute into the projection formula: \[ \text{proj}_{\text{span}(\mathbf{v}_1, \mathbf{v}_2)} (9\mathbf{e}_1) = \frac{18}{9} \mathbf{v}_1 + \frac{-18}{9} \mathbf{v}_2 = 2\mathbf{v}_1 - 2\mathbf{v}_2. \] #### Compute \( 2\mathbf{v}_1 - 2\mathbf{v}_2 \): \[ 2\mathbf{v}_1 = 2[2, 2, 1, 0] = [4, 4, 2, 0], \] \[ -2\mathbf{v}_2 = -2[-2, 2, 0, 1] = [4, -4, 0, -2], \] \[ 2\mathbf{v}_1 - 2\mathbf{v}_2 = [4, 4, 2, 0] + [4, -4, 0, -2] = [8, 0, 2, -2]. \] ### Final Answer: The orthogonal projection of \( 9\mathbf{e}_1 \) onto the subspace is: \[ [8, 0, 2, -2]. \] |
[8, 0, 2, -2] |