Math Problem Statement

Exercise 1 Determine whether the following sets are subspaces of \mathbb{R} ^ 3 under the operations of addition and scalar multiplication defined on \mathbb{R} ^ 3 Justify your answers. W_{1} = {[[x], [y], [z]] \in \mathbb{R} ^ 3, x = 3y, z = - y} 2) W2 = y ER3,2x-7y+z = 0 2. } Z (3) W_{3} = {[[x], [y], [z]] \in \mathbb{R} ^ 3, x ^ 2 - 3z ^ 2 = 0}

Solution

To determine if each set W1W_1, W2W_2, and W3W_3 is a subspace of R3\mathbb{R}^3, we need to check three criteria:

  1. The zero vector 0\mathbf{0} is in the set.
  2. The set is closed under vector addition.
  3. The set is closed under scalar multiplication.

Let's analyze each set:

1. W1={[x,y,z]R3x=3y,z=y}W_1 = \{ [x, y, z] \in \mathbb{R}^3 \mid x = 3y, z = -y \}

Zero vector:

  • Check if [0,0,0][0, 0, 0] is in W1W_1:
    • For y=0y = 0, x=3(0)=0x = 3(0) = 0 and z=0=0z = -0 = 0. Thus, [0,0,0]W1[0, 0, 0] \in W_1.

Closure under addition:

  • Take two vectors [3y1,y1,y1][3y_1, y_1, -y_1] and [3y2,y2,y2][3y_2, y_2, -y_2]: [ [3y_1, y_1, -y_1] + [3y_2, y_2, -y_2] = [3y_1 + 3y_2, y_1 + y_2, -y_1 - y_2] ] Set y=y1+y2y = y_1 + y_2, then: x=3y=3(y1+y2),z=y=(y1+y2)x = 3y = 3(y_1 + y_2), \quad z = -y = -(y_1 + y_2) Hence, the result is in W1W_1.

Closure under scalar multiplication:

  • For scalar cc and vector [3y,y,y][3y, y, -y]: [ c[3y, y, -y] = [3(cy), cy, -cy] ] Set y=cyy' = cy, then: x=3yandz=yx = 3y' \quad \text{and} \quad z = -y' Thus, the result is in W1W_1.

Conclusion:

  • W1W_1 is a subspace of R3\mathbb{R}^3.

2. W2={[x,y,z]R32x7y+z=0}W_2 = \{ [x, y, z] \in \mathbb{R}^3 \mid 2x - 7y + z = 0 \}

Zero vector:

  • Check [0,0,0][0, 0, 0]: 2(0)7(0)+0=0[0,0,0]W2.2(0) - 7(0) + 0 = 0 \quad \Rightarrow \quad [0, 0, 0] \in W_2.

Closure under addition:

  • Take [x1,y1,z1][x_1, y_1, z_1] and [x2,y2,z2][x_2, y_2, z_2]: 2x17y1+z1=0and2x27y2+z2=02x_1 - 7y_1 + z_1 = 0 \quad \text{and} \quad 2x_2 - 7y_2 + z_2 = 0 Adding gives: 2(x1+x2)7(y1+y2)+(z1+z2)=02(x_1 + x_2) - 7(y_1 + y_2) + (z_1 + z_2) = 0 Thus, closure under addition holds.

Closure under scalar multiplication:

  • For scalar cc: 2(cx)7(cy)+(cz)=c(2x7y+z)=02(cx) - 7(cy) + (cz) = c(2x - 7y + z) = 0 Hence, closure under scalar multiplication holds.

Conclusion:

  • W2W_2 is a subspace of R3\mathbb{R}^3.

3. W3={[x,y,z]R3x23z2=0}W_3 = \{ [x, y, z] \in \mathbb{R}^3 \mid x^2 - 3z^2 = 0 \}

Zero vector:

  • Check [0,0,0][0, 0, 0]: 023(0)2=0[0,0,0]W3.0^2 - 3(0)^2 = 0 \quad \Rightarrow \quad [0, 0, 0] \in W_3.

Closure under addition:

  • Take [x1,y1,z1][x_1, y_1, z_1] and [x2,y2,z2][x_2, y_2, z_2]: x123z12=0andx223z22=0x_1^2 - 3z_1^2 = 0 \quad \text{and} \quad x_2^2 - 3z_2^2 = 0 This implies: x12=3z12andx22=3z22.x_1^2 = 3z_1^2 \quad \text{and} \quad x_2^2 = 3z_2^2. However, for [x1+x2,y1+y2,z1+z2][x_1 + x_2, y_1 + y_2, z_1 + z_2]: (x1+x2)23(z1+z2)20 in general.(x_1 + x_2)^2 - 3(z_1 + z_2)^2 \neq 0 \text{ in general}. Thus, it may not be in W3W_3.

Conclusion:

  • W3W_3 is not a subspace of R3\mathbb{R}^3.

Summary

  • W1W_1: Subspace
  • W2W_2: Subspace
  • W3W_3: Not a subspace

Do you want further details on any of these points or have any questions?

Here are some follow-up questions:

  1. How do you prove that a set is not a subspace?
  2. What are other examples of subspaces in R3\mathbb{R}^3?
  3. Can you explain the geometric interpretation of subspaces?
  4. What conditions must be satisfied for a set to be closed under addition?
  5. How do you find the dimension of a subspace?

Tip: Always verify the zero vector first when checking for subspaces; it’s a key indicator.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces

Formulas

Closure under addition
Closure under scalar multiplication

Theorems

Subspace Theorem

Suitable Grade Level

Grades 11-12