| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if W = {[x, y] in R^2: x >= 0 and y >= 0} is a subspace of R^2, we need to check if W satisfies the three subspace properties: 1. The zero vector is in W. 2. W is closed under vector addition. 3. W is closed under scalar multiplication. Let's check these properties one by one: 1. The zero vector is in W. The zero vector in R^2 is [0, 0]. Since 0 >= 0 and 0 >= 0, the zero vector is indeed in W. 2. W is closed under vector addition. Let [x1, y1] and [x2, y2] be two vectors in W. Then x1 >= 0, y1 >= 0, x2 >= 0, and y2 >= 0. The sum of these two vectors is [x1 + x2, y1 + y2]. Since x1 >= 0, x2 >= 0, then x1 + x2 >= 0. Similarly, since y1 >= 0, y2 >= 0, then y1 + y2 >= 0. Therefore, [x1 + x2, y1 + y2] is also in W, and W is closed under vector addition. 3. W is closed under scalar multiplication. Let [x, y] be a vector in W and c be a scalar. Then x >= 0 and y >= 0. If c >= 0, then cx >= 0 and cy >= 0. If c < 0, then cx < 0 and cy < 0, which means that [cx, cy] is not in W. Therefore, W is not closed under scalar multiplication when the scalar is negative. Since W is not closed under scalar multiplication for all scalars, W is not a subspace of R^2. |
False |
| deepseekr1 | To determine if \( W = \{[x, y] \in \mathbb{R}^2 : x \geq 0 \text{ and } y \geq 0\} \) is a subspace of \( \mathbb{R}^2 \), we verify the three subspace criteria: 1. **Zero Vector**: \( [0, 0] \in W \), since \( 0 \geq 0 \). ✔️ 2. **Closed Under Addition**: For \( [x_1, y_1], [x_2, y_2] \in W \), their sum \( [x_1 + x_2, y_1 + y_2] \) has \( x_1 + x_2 \geq 0 \) and \( y_1 + y_2 \geq 0 \), so it remains in \( W \). ✔️ 3. **Closed Under Scalar Multiplication**: Let \( \alpha \in \mathbb{R} \). If \( \alpha < 0 \) (e.g., \( \alpha = -1 \)) and \( [x, y] \in W \), then \( \alpha[x, y] = [\alpha x, \alpha y] \) has negative components, which are not in \( W \). ❌ **Conclusion**: Since \( W \) is not closed under scalar multiplication, it is **not** a subspace of \( \mathbb{R}^2 \). **Answer**: No, \( W \) is not a subspace of \( \mathbb{R}^2 \) because it fails closure under scalar multiplication. |
False |
| gpto1 | To determine whether the set \( W = \{ [x, y] \in \mathbb{R}^2 : x \geq 0 \text{ and } y \geq 0 \} \) is a subspace of \( \mathbb{R}^2 \), we'll verify if it satisfies the three key properties that define a subspace: 1. **Non-emptiness (Contains the zero vector)** 2. **Closure under vector addition** 3. **Closure under scalar multiplication** ### 1. Non-emptiness A subspace must contain the zero vector. - The zero vector in \( \mathbb{R}^2 \) is \( \mathbf{0} = [0, 0] \). - Since \( 0 \geq 0 \) and \( 0 \geq 0 \), \( \mathbf{0} \in W \). **Conclusion:** \( W \) is non-empty as it contains the zero vector. ### 2. Closure under Vector Addition For any vectors \( \mathbf{u}, \mathbf{v} \in W \), their sum \( \mathbf{u} + \mathbf{v} \) must also be in \( W \). Let \( \mathbf{u} = [x_1, y_1] \) and \( \mathbf{v} = [x_2, y_2] \) where \( x_1, x_2, y_1, y_2 \geq 0 \). - Sum: \( \mathbf{u} + \mathbf{v} = [x_1 + x_2, y_1 + y_2] \) - Since \( x_1, x_2 \geq 0 \), their sum \( x_1 + x_2 \geq 0 \). - Similarly, \( y_1 + y_2 \geq 0 \). **Conclusion:** \( \mathbf{u} + \mathbf{v} \in W \), so \( W \) is closed under addition. ### 3. Closure under Scalar Multiplication For any vector \( \mathbf{u} \in W \) and any scalar \( c \in \mathbb{R} \), the product \( c\mathbf{u} \) must also be in \( W \). Let \( \mathbf{u} = [x, y] \) where \( x, y \geq 0 \). Consider scalar \( c \in \mathbb{R} \): #### Case 1: \( c \geq 0 \) - \( c x \geq 0 \) since \( c \geq 0 \) and \( x \geq 0 \). - \( c y \geq 0 \) since \( c \geq 0 \) and \( y \geq 0 \). - Therefore, \( c\mathbf{u} = [c x, c y] \in W \). #### Case 2: \( c < 0 \) - \( c x \leq 0 \) since \( c < 0 \) and \( x \geq 0 \). - \( c y \leq 0 \) since \( c < 0 \) and \( y \geq 0 \). - Therefore, at least one of \( c x \) or \( c y \) is not greater than or equal to zero. **Example:** Let \( \mathbf{u} = [1, 1] \in W \) and \( c = -1 \). - \( c\mathbf{u} = -1 \cdot [1, 1] = [-1, -1] \). - Since \( -1 \not\geq 0 \), \( c\mathbf{u} \notin W \). **Conclusion:** \( W \) is **not** closed under scalar multiplication when \( c < 0 \). ### Final Conclusion Since \( W \) fails to satisfy closure under scalar multiplication for all real scalars, it is **not** a subspace of \( \mathbb{R}^2 \). --- **Answer:** No, \( W \) is not a subspace of \( \mathbb{R}^2 \) because it is not closed under scalar multiplication by negative scalars. |
False |
| GPT-4o | To determine whether the set \( W = \{ [x, y] \in \mathbb{R}^2 : x \geq 0 \text{ and } y \geq 0 \} \) is a subspace of \(\mathbb{R}^2\), we need to check if \( W \) satisfies the three main criteria for being a subspace: 1. **Zero Vector**: The zero vector \( \mathbf{0} = [0, 0] \) must be in \( W \). 2. **Closed under Addition**: If \( \mathbf{u} \) and \( \mathbf{v} \) are in \( W \), then \( \mathbf{u} + \mathbf{v} \) must also be in \( W \). 3. **Closed under Scalar Multiplication**: If \( \mathbf{u} \) is in \( W \) and \( c \) is a scalar, then \( c \mathbf{u} \) must also be in \( W \). Let's check each condition: 1. **Zero Vector**: The zero vector \( [0, 0] \) is in \( W \) because both \( 0 \geq 0 \) and \( 0 \geq 0 \) are true. So, this condition is satisfied. 2. **Closed under Addition**: Let \( \mathbf{u} = [x_1, y_1] \in W \) and \( \mathbf{v} = [x_2, y_2] \in W \). This means \( x_1 \geq 0 \), \( y_1 \geq 0 \), \( x_2 \geq 0 \), and \( y_2 \geq 0 \). Consider \( \mathbf{u} + \mathbf{v} = [x_1 + x_2, y_1 + y_2] \). Since \( x_1 + x_2 \geq 0 \) and \( y_1 + y_2 \geq 0 \), \( \mathbf{u} + \mathbf{v} \) is also in \( W \). Therefore, \( W \) is closed under addition. 3. **Closed under Scalar Multiplication**: Let \( \mathbf{u} = [x, y] \in W \) and let \( c \) be a scalar. We need to consider two cases: - If \( c \geq 0 \), then \( c x \geq 0 \) and \( c y \geq 0 \), so \( c \mathbf{u} = [c x, c y] \in W \). - If \( c < 0 \), then \( c x \leq 0 \) and \( c y \leq 0 \). For \( \mathbf{u} \) to remain in \( W \), both \( c x \geq 0 \) and \( c y \geq 0 \) must hold, which cannot be true if \( c < 0 \). Hence, \( c \mathbf{u} \) is not necessarily in \( W \) when \( c < 0 \). Since \( W \) fails the scalar multiplication condition when \( c \) is negative, \( W \) is not closed under scalar multiplication for all scalars. **Conclusion**: \( W \) is not a subspace of \(\mathbb{R}^2\) because it does not satisfy the requirement of being closed under scalar multiplication for all real numbers. |
False |
| llama3.1-70B-Instruct | To determine if W = {[x, y] in R^2: x >= 0 and y >= 0} is a subspace of R^2, we need to check if it satisfies the three properties of a subspace: 1. **Closure under addition**: For any two vectors [x1, y1] and [x2, y2] in W, their sum [x1 + x2, y1 + y2] must also be in W. Let's take two vectors [x1, y1] and [x2, y2] in W, where x1 >= 0, y1 >= 0, x2 >= 0, and y2 >= 0. Then, their sum is: [x1 + x2, y1 + y2] Since x1 + x2 >= 0 and y1 + y2 >= 0 (because the sum of non-negative numbers is non-negative), the sum [x1 + x2, y1 + y2] is also in W. 2. **Closure under scalar multiplication**: For any vector [x, y] in W and any scalar c in R, the product c[x, y] = [cx, cy] must also be in W. Let's take a vector [x, y] in W, where x >= 0 and y >= 0, and a scalar c in R. Then, the product is: [cx, cy] If c >= 0, then cx >= 0 and cy >= 0, so [cx, cy] is in W. However, if c < 0, then cx < 0 and cy < 0, so [cx, cy] is not in W. Therefore, W is not closed under scalar multiplication. Since W fails to satisfy one of the properties of a subspace, we can conclude that: W = {[x, y] in R^2: x >= 0 and y >= 0} is not a subspace of R^2. |
False |
| llama3.1-405B-Instruct | To determine if W = {[x, y] in R^2: x >= 0 and y >= 0} is a subspace of R^2, we need to check if it satisfies the three properties of a subspace: 1. W is non-empty: This is true, since the zero vector [0, 0] is in W. 2. W is closed under addition: Let [x1, y1] and [x2, y2] be two vectors in W. Then, x1 >= 0, y1 >= 0, x2 >= 0, and y2 >= 0. Adding these vectors, we get [x1 + x2, y1 + y2]. Since x1 + x2 >= 0 and y1 + y2 >= 0, the sum is also in W. 3. W is closed under scalar multiplication: Let [x, y] be a vector in W and c be a scalar. Then, x >= 0 and y >= 0. Multiplying [x, y] by c, we get [cx, cy]. Since cx >= 0 and cy >= 0, the scalar multiple is also in W. Since W satisfies all three properties, it is a subspace of R^2. |
False |
| DeepSeek-Coder-V2 | To determine whether the set W = {[x, y] in R^2: x >= 0 and y >= 0} is a subspace of R^2, we need to check if it satisfies the three subspace properties: 1. **Contains the zero vector**: The zero vector in R^2 is [0, 0]. Since 0 >= 0 and 0 >= 0, the zero vector [0, 0] is in W. Therefore, W contains the zero vector. 2. **Closed under vector addition**: For any two vectors [x1, y1] and [x2, y2] in W, their sum [x1 + x2, y1 + y2] must also be in W. Since x1 >= 0, x2 >= 0, y1 >= 0, and y2 >= 0, it follows that x1 + x2 >= 0 and y1 + y2 >= 0. Therefore, the sum [x1 + x2, y1 + y2] is in W. Thus, W is closed under vector addition. 3. **Closed under scalar multiplication**: For any vector [x, y] in W and any scalar c in R, the product c[x, y] = [cx, cy] must also be in W. If c is non-negative (c >= 0), then cx >= 0 and cy >= 0, so [cx, cy] is in W. However, if c is negative (c < 0), then cx < 0 and cy < 0, which means [cx, cy] is not in W. Therefore, W is not closed under scalar multiplication by all real numbers. Since W is not closed under scalar multiplication by all real numbers, it does not satisfy all the subspace properties. Therefore, W is not a subspace of R^2. |
False |
| Qwen2-72B-Instruct | To determine if W = {[x, y] in R^2: x >= 0 and y >= 0} is a subspace of R^2, we need to check if W satisfies the three subspace properties: 1. The zero vector is in W. 2. W is closed under vector addition. 3. W is closed under scalar multiplication. Let's check these properties one by one: 1. The zero vector is in W. The zero vector in R^2 is [0, 0]. Since 0 >= 0 and 0 >= 0, the zero vector is indeed in W. 2. W is closed under vector addition. Let [x1, y1] and [x2, y2] be two vectors in W. Then x1 >= 0, y1 >= 0, x2 >= 0, and y2 >= 0. The sum of these two vectors is [x1 + x2, y1 + y2]. Since x1 >= 0, x2 >= 0, then x1 + x2 >= 0. Similarly, since y1 >= 0, y2 >= 0, then y1 + y2 >= 0. Therefore, [x1 + x2, y1 + y2] is also in W, and W is closed under vector addition. 3. W is closed under scalar multiplication. Let [x, y] be a vector in W and c be a scalar. Then x >= 0 and y >= 0. If c >= 0, then cx >= 0 and cy >= 0. If c < 0, then cx < 0 and cy < 0, which means that [cx, cy] is not in W. Therefore, W is not closed under scalar multiplication when the scalar is negative. Since W is not closed under scalar multiplication for all scalars, W is not a subspace of R^2. |
False |
| baichuan4 | To determine if W = {[x, y] ∈ R^2: x ≥ 0 and y ≥ 0} is a subspace of R^2, we need to check if it satisfies the three properties of a subspace: 1. The zero vector (0, 0) must be in W. 2. For any vectors u = [u1, u2] and v = [v1, v2] in W, the sum u + v must also be in W. 3. For any vector u = [u1, u2] in W and any scalar c ∈ R, the product cu must also be in W. Let's check these properties: 1. The zero vector (0, 0) is in W because both its components are non-negative. 2. Let u = [u1, u2] and v = [v1, v2] be vectors in W. Then, u + v = [u1 + v1, u2 + v2]. Since u1, u2, v1, and v2 are all non-negative, their sums u1 + v1 and u2 + v2 will also be non-negative. Therefore, u + v is in W. 3. Let u = [u1, u2] be a vector in W and let c be any scalar in R. Then, cu = [cu1, cu2]. If c is positive or zero, then cu1 and cu2 will be non-negative, so cu will be in W. However, if c is negative, then cu1 and cu2 will be non-positive, which means cu will not be in W. Since W does not satisfy the third property of a subspace when c is negative, W is not a subspace of R^2. |
False |
| doubao | To determine if W is a subspace of R^2, we need to check three conditions: 1. The zero vector is in W. The zero vector in R^2 is [0, 0], and since both 0 >= 0, the zero vector is in W. 2. Closure under addition. Let [x1, y1] and [x2, y2] be in W. Then x1 >= 0 and y1 >= 0, and x2 >= 0 and y2 >= 0. Their sum is [x1 + x2, y1 + y2]. Since x1 + x2 >= 0 and y1 + y2 >= 0, the sum is in W, so W is closed under addition. 3. Closure under scalar multiplication. Let [x, y] be in W and c be a scalar. Then x >= 0 and y >= 0. c[x, y] = [cx, cy]. Since cx >= 0 and cy >= 0 for a non-negative scalar c, W is closed under scalar multiplication. Since all three conditions are satisfied, W is a subspace of R^2. |
False |
| minimax | To determine whether a set W is a subspace of R^2, we need to check if it satisfies the three subspace properties: 1. The zero vector is in W. 2. W is closed under vector addition. 3. W is closed under scalar multiplication. Let's check each property for the set W = {[x, y] in R^2: x >= 0 and y >= 0}: 1. Zero vector in W: The zero vector in R^2 is [0, 0]. Since both components of the zero vector are non-negative (in fact, they are zero), the zero vector is indeed in W. So, W satisfies the first property. 2. Closure under vector addition: Let [x1, y1] and [x2, y2] be two vectors in W. This means x1 >= 0, y1 >= 0, x2 >= 0, and y2 >= 0. When we add these two vectors, we get [x1 + x2, y1 + y2]. Since the sum of two non-negative numbers is non-negative, x1 + x2 >= 0 and y1 + y2 >= 0. Therefore, the resulting vector is also in W. So, W satisfies the second property. 3. Closure under scalar multiplication: Let [x, y] be a vector in W, which means x >= 0 and y >= 0. Now, consider a scalar k in R. If we multiply the vector [x, y] by the scalar k, we get [kx, ky]. If k is non-negative, then kx >= 0 and ky >= 0, and the resulting vector is in W. However, if k is negative, then kx and ky could be negative, which would result in a vector that is not in W. Therefore, W is not closed under scalar multiplication for all possible scalars in R. Since W fails the third property (closure under scalar multiplication for all scalars), it is not a subspace of R^2. |
False |
| kimi | To determine if W is a subspace of R^2, we need to check if it satisfies the three subspace criteria: 1. The zero vector (0,0) is in W. 2. W is closed under vector addition. 3. W is closed under scalar multiplication. Let's check each criterion: 1. The zero vector (0,0) is in W because 0 >= 0 and 0 >= 0. 2. W is closed under vector addition: Let [x1, y1] and [x2, y2] be vectors in W. Then x1 >= 0, y1 >= 0, x2 >= 0, and y2 >= 0. When we add these vectors, we get [x1 + x2, y1 + y2]. Since x1 >= 0 and x2 >= 0, their sum x1 + x2 is also greater than or equal to 0. Similarly, y1 >= 0 and y2 >= 0, so y1 + y2 is also greater than or equal to 0. Therefore, [x1 + x2, y1 + y2] is also in W. 3. W is closed under scalar multiplication: Let [x, y] be a vector in W and let c be a scalar. Then x >= 0 and y >= 0. When we multiply the vector by the scalar, we get [cx, cy]. If c >= 0, then cx >= 0 and cy >= 0, so [cx, cy] is in W. If c < 0, then cx < 0 and/or cy < 0, so [cx, cy] is not in W. Since W is not closed under scalar multiplication for all scalars (specifically, it is not closed for negative scalars), W is not a subspace of R^2. Solution: W is not a subspace of R^2. |
False |
| wenxinyiyan | 首先，我们需要明确子空间的定义。在R^2中，一个子空间需要满足以下三个条件： 1. 包含原点：子空间中必须包含原点（0,0）。 2. 封闭性：对于子空间中的任意两个元素，它们的和仍然在子空间中。 3. 数乘封闭性：对于子空间中的任意元素和任意实数，它们的乘积仍然在子空间中。 现在，我们来判断W是否满足这些条件。 1. 包含原点：由于x >= 0 and y >= 0，原点（0,0）显然在W中。 2. 封闭性：考虑W中的两个元素[x1, y1]和[x2, y2]，它们的和为[x1+x2, y1+y2]。由于x1, x2, y1, y2都大于等于0，所以x1+x2和y1+y2也都大于等于0。因此，[x1+x2, y1+y2]也在W中。 3. 数乘封闭性：考虑W中的元素[x, y]和任意实数k，它们的乘积为[kx, ky]。如果k是非负的，那么kx和ky都大于等于0，所以[kx, ky]在W中。但是，如果k是负数，那么kx和ky可能小于0，所以[kx, ky]可能不在W中。 由于W不满足数乘封闭性，所以W不是R^2的子空间。 |
False |
| zhipu | To determine if \( W = \{ [x, y] \in \mathbb{R}^2 : x \geq 0 \text{ and } y \geq 0 \} \) is a subspace of \( \mathbb{R}^2 \), we need to check if \( W \) satisfies the three criteria for being a subspace: 1. **Contains the Zero Vector:** The zero vector \( [0, 0] \) must be in \( W \). 2. **Closed under Vector Addition:** If \( \mathbf{u} \) and \( \mathbf{v} \) are in \( W \), then \( \mathbf{u} + \mathbf{v} \) must also be in \( W \). 3. **Closed under Scalar Multiplication:** If \( \mathbf{u} \) is in \( W \) and \( c \) is any scalar, then \( c\mathbf{u} \) must also be in \( W \). ### Step-by-Step Verification: 1. **Contains the Zero Vector:** - The zero vector in \( \mathbb{R}^2 \) is \( [0, 0] \). - Clearly, \( [0, 0] \in W \) because \( 0 \geq 0 \) and \( 0 \geq 0 \). 2. **Closed under Vector Addition:** - Let \( \mathbf{u} = [x_1, y_1] \) and \( \mathbf{v} = [x_2, y_2] \) be vectors in \( W \). - This means \( x_1 \geq 0 \), \( y_1 \geq 0 \), \( x_2 \geq 0 \), and \( y_2 \geq 0 \). - Consider the vector sum \( \mathbf{u} + \mathbf{v} = [x_1 + x_2, y_1 + y_2] \). - Since \( x_1 \geq 0 \) and \( x_2 \geq 0 \), it follows that \( x_1 + x_2 \geq 0 \). - Similarly, since \( y_1 \geq 0 \) and \( y_2 \geq 0 \), it follows that \( y_1 + y_2 \geq 0 \). - Therefore, \( \mathbf{u} + \mathbf{v} \in W \). 3. **Closed under Scalar Multiplication:** - Let \( \mathbf{u} = [x, y] \) be a vector in \( W \) and \( c \) be any scalar. - This means \( x \geq 0 \) and \( y \geq 0 \). - Consider the scalar multiple \( c\mathbf{u} = [cx, cy] \). - For \( c\mathbf{u} \) to be in \( W \), both \( cx \) and \( cy \) must be non-negative. - However, if \( c \) is a negative scalar, then \( cx \) and \( cy \) can be negative, which means \( c\mathbf{u} \notin W \). ### Conclusion: While \( W \) contains the zero vector and is closed under vector addition, it is **not** closed under scalar multiplication when the scalar is negative. Therefore, \( W \) does not satisfy all the criteria to be a subspace of \( \mathbb{R}^2 \). **Answer: \( W \) is not a subspace of \( \mathbb{R}^2 \).** |
False |
| gpto1 | To determine whether the set \( W = \{ [x, y] \in \mathbb{R}^2 : x \geq 0 \text{ and } y \geq 0 \} \) is a subspace of \( \mathbb{R}^2 \), we'll verify if it satisfies the three key properties that define a subspace: 1. **Non-emptiness (Contains the zero vector)** 2. **Closure under vector addition** 3. **Closure under scalar multiplication** ### 1. Non-emptiness A subspace must contain the zero vector. - The zero vector in \( \mathbb{R}^2 \) is \( \mathbf{0} = [0, 0] \). - Since \( 0 \geq 0 \) and \( 0 \geq 0 \), \( \mathbf{0} \in W \). **Conclusion:** \( W \) is non-empty as it contains the zero vector. ### 2. Closure under Vector Addition For any vectors \( \mathbf{u}, \mathbf{v} \in W \), their sum \( \mathbf{u} + \mathbf{v} \) must also be in \( W \). Let \( \mathbf{u} = [x_1, y_1] \) and \( \mathbf{v} = [x_2, y_2] \) where \( x_1, x_2, y_1, y_2 \geq 0 \). - Sum: \( \mathbf{u} + \mathbf{v} = [x_1 + x_2, y_1 + y_2] \) - Since \( x_1, x_2 \geq 0 \), their sum \( x_1 + x_2 \geq 0 \). - Similarly, \( y_1 + y_2 \geq 0 \). **Conclusion:** \( \mathbf{u} + \mathbf{v} \in W \), so \( W \) is closed under addition. ### 3. Closure under Scalar Multiplication For any vector \( \mathbf{u} \in W \) and any scalar \( c \in \mathbb{R} \), the product \( c\mathbf{u} \) must also be in \( W \). Let \( \mathbf{u} = [x, y] \) where \( x, y \geq 0 \). Consider scalar \( c \in \mathbb{R} \): #### Case 1: \( c \geq 0 \) - \( c x \geq 0 \) since \( c \geq 0 \) and \( x \geq 0 \). - \( c y \geq 0 \) since \( c \geq 0 \) and \( y \geq 0 \). - Therefore, \( c\mathbf{u} = [c x, c y] \in W \). #### Case 2: \( c < 0 \) - \( c x \leq 0 \) since \( c < 0 \) and \( x \geq 0 \). - \( c y \leq 0 \) since \( c < 0 \) and \( y \geq 0 \). - Therefore, at least one of \( c x \) or \( c y \) is not greater than or equal to zero. **Example:** Let \( \mathbf{u} = [1, 1] \in W \) and \( c = -1 \). - \( c\mathbf{u} = -1 \cdot [1, 1] = [-1, -1] \). - Since \( -1 \not\geq 0 \), \( c\mathbf{u} \notin W \). **Conclusion:** \( W \) is **not** closed under scalar multiplication when \( c < 0 \). ### Final Conclusion Since \( W \) fails to satisfy closure under scalar multiplication for all real scalars, it is **not** a subspace of \( \mathbb{R}^2 \). --- **Answer:** No, \( W \) is not a subspace of \( \mathbb{R}^2 \) because it is not closed under scalar multiplication by negative scalars. |
False |
| deepseekv3 | To determine whether \( W = \{[x, y] \in \mathbb{R}^2: x \geq 0 \text{ and } y \geq 0\} \) is a subspace of \( \mathbb{R}^2 \), we need to verify if \( W \) satisfies the three properties of a subspace: 1. **Contains the zero vector**: The zero vector \( [0, 0] \) must be in \( W \). 2. **Closed under vector addition**: If \( \mathbf{u} = [u_1, u_2] \) and \( \mathbf{v} = [v_1, v_2] \) are in \( W \), then \( \mathbf{u} + \mathbf{v} \) must also be in \( W \). 3. **Closed under scalar multiplication**: If \( \mathbf{u} = [u_1, u_2] \) is in \( W \) and \( c \) is a scalar, then \( c\mathbf{u} \) must also be in \( W \). --- ### Step 1: Contains the zero vector The zero vector \( [0, 0] \) satisfies \( x \geq 0 \) and \( y \geq 0 \), so \( [0, 0] \in W \). **Property 1 is satisfied.** --- ### Step 2: Closed under vector addition Let \( \mathbf{u} = [u_1, u_2] \) and \( \mathbf{v} = [v_1, v_2] \) be in \( W \), so \( u_1, u_2, v_1, v_2 \geq 0 \). Then, \( \mathbf{u} + \mathbf{v} = [u_1 + v_1, u_2 + v_2] \). Since \( u_1 + v_1 \geq 0 \) and \( u_2 + v_2 \geq 0 \), \( \mathbf{u} + \mathbf{v} \in W \). **Property 2 is satisfied.** --- ### Step 3: Closed under scalar multiplication Let \( \mathbf{u} = [u_1, u_2] \in W \), so \( u_1, u_2 \geq 0 \). For scalar multiplication, consider \( c \in \mathbb{R} \). If \( c \geq 0 \), then \( c\mathbf{u} = [cu_1, cu_2] \in W \) because \( cu_1, cu_2 \geq 0 \). However, if \( c < 0 \), then \( cu_1, cu_2 \leq 0 \), so \( c\mathbf{u} \notin W \). Since \( W \) is not closed under scalar multiplication for all scalars, **Property 3 is not satisfied.** --- ### Conclusion Since \( W \) fails to satisfy the third property of a subspace, \( W \) is **not a subspace of \( \mathbb{R}^2 \)**. |
False |