A set in R² is displayed to the right. Assume the set includes the bounding lines. Give a specific reason why the set H is not a subspace of R². (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.) Let u and v be vectors and let k be a scalar. Select the correct choice below and, if necessary, fill in the answer box within your choice. OA. The set is not a subspace because it is closed under sums, but not under scalar multiplication. For example, multiplied by (0,1) is not in the set. Q G O B. The set is not a subspace because it is not closed under either scalar multiplication or sums. For example, multiplied by (0.1) is not in the set, and the sum of (2,2) and (-1,-3) is not in the set. Q U+V ku OC. The set is not a subspace because it is closed under scalar multiplication, but not under sums. For example, the sum of (2.2) and (-1,-3) is not in the set. Q Q Q Q D. The set is not a subspace because it does not include the zero vector.
A set in R² is displayed to the right. Assume the set includes the bounding lines. Give a specific reason why the set H is not a subspace of R². (For instance, find two vectors in H whose sum is not in H, or find a vector in H with a scalar multiple that is not in H. Draw a picture.) Let u and v be vectors and let k be a scalar. Select the correct choice below and, if necessary, fill in the answer box within your choice. OA. The set is not a subspace because it is closed under sums, but not under scalar multiplication. For example, multiplied by (0,1) is not in the set. Q G O B. The set is not a subspace because it is not closed under either scalar multiplication or sums. For example, multiplied by (0.1) is not in the set, and the sum of (2,2) and (-1,-3) is not in the set. Q U+V ku OC. The set is not a subspace because it is closed under scalar multiplication, but not under sums. For example, the sum of (2.2) and (-1,-3) is not in the set. Q Q Q Q D. The set is not a subspace because it does not include the zero vector.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
*Illustration of a set in \( \mathbb{R}^2 \), bounded by lines.*
#### Explanation Options
Let \( u \) and \( v \) be vectors, and let \( k \) be a scalar.
- **Option A**: The set is not a subspace because it is closed under sums, but not under scalar multiplication.
- *Example*: The vector \( (0,1) \) is in the set, however \( 0 \cdot (0,1) \) is not in the set.
- *Illustration*:
- **Option B**: The set is not a subspace because it is not closed under either scalar multiplication or sums.
- *Example*: The vector \( (2,2) \) is multiplied by \( 0 \), which is not \( (0,1) \). Also, the sum of \( (2,2) \) and \( (-1,-3) \) is not in the set.
- *Illustration*:
- **Option C**: The set is not a subspace because it is closed under scalar multiplication, but not under sums.
- *Example*: The sum of the vectors \( (2,2) \) and \( (-1,-3) \) is not in the set.
- *Illustration*:
- **Option D**: The set is not a subspace because it does not include the zero vector.
- *Example*: The zero vector \((0,0)\) is not](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcbfb1871-07e2-4dce-8005-269462079c4a%2Ff3a4cb29-7592-480d-bdf9-4ef493a867f1%2F3wl07v5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Assessing Subspaces in \( \mathbb{R}^2 \)
In linear algebra, understanding the concept of subspaces is essential. Below is a discussion focused on determining why a particular set in \( \mathbb{R}^2 \) is not a subspace of \( \mathbb{R}^2 \).
#### Problem Statement
A set in \( \mathbb{R}^2 \) is displayed. Assume the set includes the bounding lines. You're required to give a specific reason why the set \( H \) is not a subspace of \( \mathbb{R}^2 \).
*Illustration of a set in \( \mathbb{R}^2 \), bounded by lines.*
#### Explanation Options
Let \( u \) and \( v \) be vectors, and let \( k \) be a scalar.
- **Option A**: The set is not a subspace because it is closed under sums, but not under scalar multiplication.
- *Example*: The vector \( (0,1) \) is in the set, however \( 0 \cdot (0,1) \) is not in the set.
- *Illustration*:
- **Option B**: The set is not a subspace because it is not closed under either scalar multiplication or sums.
- *Example*: The vector \( (2,2) \) is multiplied by \( 0 \), which is not \( (0,1) \). Also, the sum of \( (2,2) \) and \( (-1,-3) \) is not in the set.
- *Illustration*:
- **Option C**: The set is not a subspace because it is closed under scalar multiplication, but not under sums.
- *Example*: The sum of the vectors \( (2,2) \) and \( (-1,-3) \) is not in the set.
- *Illustration*:
- **Option D**: The set is not a subspace because it does not include the zero vector.
- *Example*: The zero vector \((0,0)\) is not
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