https://file.aiquickdraw.com/mathbot/file/1726983417053rcs8wfj6.jpg

Let's go through each part of the problem carefully.

### Problem 7:
Let \( V \) be the first quadrant in the \( xy \)-plane, i.e., 
\[
V = \left\{ \begin{pmatrix} x \\ y \end{pmatrix} \mid x \geq 0, y \geq 0 \right\}.
\]
This defines the set of all vectors in the first quadrant of the plane.

### Part (a):
*If \( \mathbf{u} \) and \( \mathbf{v} \) are in \( V \), is \( \mathbf{u} + \mathbf{v} \) in \( V \)? Why?*

#### Solution:
- Let \( \mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} \) and \( \mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} \) be two vectors in \( V \).
- This means that \( u_1 \geq 0 \), \( u_2 \geq 0 \) and \( v_1 \geq 0 \), \( v_2 \geq 0 \).
- The sum of the vectors \( \mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix} \).
  
Since both \( u_1 \geq 0 \) and \( v_1 \geq 0 \), we know \( u_1 + v_1 \geq 0 \). Similarly, \( u_2 + v_2 \geq 0 \). Therefore, \( \mathbf{u} + \mathbf{v} \) is also in the first quadrant, which means it is in \( V \).

**Conclusion**: Yes, if \( \mathbf{u} \) and \( \mathbf{v} \) are in \( V \), then \( \mathbf{u} + \mathbf{v} \) is also in \( V \).

---

### Part (b):
*Show that \( V \) is not a vector space.*

#### Solution:
A set of vectors forms a **vector space** if it satisfies the following conditions, among others:
- Closure under addition (which we verified in part (a)).
- Closure under scalar multiplication, meaning if \( \mathbf{u} \in V \) and \( c \in \mathbb{R} \), then \( c\mathbf{u} \in V \).

Let's test closure under scalar multiplication:
- Let \( \mathbf{u} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \in V \) (this vector is in the first quadrant since both components are positive).
- Now, take a scalar \( c = -1 \). Then \( c\mathbf{u} = -1 \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix} \).

This vector \( \begin{pmatrix} -1 \\ -1 \end{pmatrix} \) is **not** in \( V \) because both components are negative (it lies in the third quadrant). Hence, \( V \) is not closed under scalar multiplication.

**Conclusion**: \( V \) is not a vector space because it is not closed under scalar multiplication. Specifically, multiplying a vector in \( V \) by a negative scalar results in a vector that is not in \( V \).

---

Let me know if you'd like further details on any part of the solution. Here are some related questions to consider:

1. Can you define other conditions necessary for a set to be a vector space?
2. How would the definition of \( V \) change if it included vectors from more quadrants of the plane?
3. What role does scalar multiplication play in defining vector spaces?
4. Is the set of all vectors in the first quadrant a subspace of \( \mathbb{R}^2 \)?
5. Could the set \( V \) form a vector space if restricted to specific operations or scalars?

**Tip**: Always check both closure under addition and scalar multiplication when verifying if a set forms a vector space.

Let V be the first quadrant in the xy-plane; that is, let V = { (x, y) | x >= 0, y >= 0 }. (a) If u and v are in V, is u + v in V? Why? (b) Show that V is not a vector space.

Explore whether the set of vectors in the first quadrant of the xy-plane forms a vector space. Analyze closure under vector addition and scalar multiplication using examples and theorems on vector space axioms.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation