Math Problem Statement

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.

Solution

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

Problem 7:

Let VV be the first quadrant in the xyxy-plane, i.e., V={(xy)x0,y0}.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 u\mathbf{u} and v\mathbf{v} are in VV, is u+v\mathbf{u} + \mathbf{v} in VV? Why?

Solution:

  • Let u=(u1u2)\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} and v=(v1v2)\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} be two vectors in VV.
  • This means that u10u_1 \geq 0, u20u_2 \geq 0 and v10v_1 \geq 0, v20v_2 \geq 0.
  • The sum of the vectors u+v=(u1+v1u2+v2)\mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix}.

Since both u10u_1 \geq 0 and v10v_1 \geq 0, we know u1+v10u_1 + v_1 \geq 0. Similarly, u2+v20u_2 + v_2 \geq 0. Therefore, u+v\mathbf{u} + \mathbf{v} is also in the first quadrant, which means it is in VV.

Conclusion: Yes, if u\mathbf{u} and v\mathbf{v} are in VV, then u+v\mathbf{u} + \mathbf{v} is also in VV.


Part (b):

Show that VV 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 uV\mathbf{u} \in V and cRc \in \mathbb{R}, then cuVc\mathbf{u} \in V.

Let's test closure under scalar multiplication:

  • Let u=(11)V\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=1c = -1. Then cu=1(11)=(11)c\mathbf{u} = -1 \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -1 \end{pmatrix}.

This vector (11)\begin{pmatrix} -1 \\ -1 \end{pmatrix} is not in VV because both components are negative (it lies in the third quadrant). Hence, VV is not closed under scalar multiplication.

Conclusion: VV is not a vector space because it is not closed under scalar multiplication. Specifically, multiplying a vector in VV by a negative scalar results in a vector that is not in VV.


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 VV 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 R2\mathbb{R}^2?
  5. Could the set VV 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.

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Vector Spaces
Closure Property
Scalar Multiplication

Formulas

u + v = (u1 + v1, u2 + v2)
c * u = (c * u1, c * u2)

Theorems

Vector space axioms
Closure under addition and scalar multiplication

Suitable Grade Level

Undergraduate Mathematics