Math Problem Statement

Find a list of four distinct vectors in 𝐅^3 whose span equals {(π‘₯, 𝑦, 𝑧) ∈ 𝐅^3 ∢ π‘₯ + 𝑦 + 𝑧 = 0}.

Solution

To find a list of four distinct vectors in F3\mathbb{F}^3 whose span equals the subspace {(x,y,z)∈F3:x+y+z=0}\{(x, y, z) \in \mathbb{F}^3 : x + y + z = 0\}, let's break down the problem step by step.

1. Identify the Subspace:

The subspace S={(x,y,z)∈F3:x+y+z=0}S = \{(x, y, z) \in \mathbb{F}^3 : x + y + z = 0\} consists of all vectors in F3\mathbb{F}^3 where the sum of the components is zero. This is a plane in F3\mathbb{F}^3, meaning it's a two-dimensional subspace of F3\mathbb{F}^3.

2. Determine the Dimension:

The equation x+y+z=0x + y + z = 0 imposes one linear constraint on the vectors in F3\mathbb{F}^3, so the dimension of this subspace is 3βˆ’1=23 - 1 = 2.

3. Find a Basis:

To find vectors that span this subspace, we need to find two linearly independent vectors v1\mathbf{v}_1 and v2\mathbf{v}_2 that satisfy the equation x+y+z=0x + y + z = 0. These vectors will form a basis for the subspace.

Two possible vectors could be: v1=(1,βˆ’1,0),v2=(0,1,βˆ’1)\mathbf{v}_1 = (1, -1, 0), \quad \mathbf{v}_2 = (0, 1, -1)

These vectors are linearly independent, and any vector in the subspace can be written as a linear combination of these two vectors.

4. Find Four Distinct Vectors:

Although the dimension of the subspace is 2, we can construct four distinct vectors by choosing appropriate linear combinations. For instance: v1=(1,βˆ’1,0)\mathbf{v}_1 = (1, -1, 0) v2=(0,1,βˆ’1)\mathbf{v}_2 = (0, 1, -1) v3=(βˆ’1,0,1)=βˆ’v1+v2\mathbf{v}_3 = (-1, 0, 1) = -\mathbf{v}_1 + \mathbf{v}_2 v4=(1,1,βˆ’2)=v1+v2\mathbf{v}_4 = (1, 1, -2) = \mathbf{v}_1 + \mathbf{v}_2

5. Verification:

The set of vectors {v1,v2,v3,v4}\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\} spans the entire subspace since any vector v\mathbf{v} in the subspace can be written as a linear combination of v1\mathbf{v}_1 and v2\mathbf{v}_2, and therefore also in terms of v3\mathbf{v}_3 and v4\mathbf{v}_4.

Final List of Vectors:

  • v1=(1,βˆ’1,0)\mathbf{v}_1 = (1, -1, 0)
  • v2=(0,1,βˆ’1)\mathbf{v}_2 = (0, 1, -1)
  • v3=(βˆ’1,0,1)\mathbf{v}_3 = (-1, 0, 1)
  • v4=(1,1,βˆ’2)\mathbf{v}_4 = (1, 1, -2)

These four distinct vectors span the subspace {(x,y,z)∈F3:x+y+z=0}\{(x, y, z) \in \mathbb{F}^3 : x + y + z = 0\}.

Would you like any details or have any questions?

Here are 5 related questions:

  1. How can we verify if a set of vectors spans a subspace?
  2. What does it mean for vectors to be linearly independent?
  3. How do we determine the dimension of a subspace?
  4. What is the geometric interpretation of a plane in F3\mathbb{F}^3?
  5. How can we find the orthogonal complement of a subspace?

Tip: When finding a spanning set for a subspace, it's helpful to start by determining the dimension of the subspace and finding a basis, which is the smallest set of vectors needed to span the space.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Vector Spaces
Subspaces

Formulas

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Theorems

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Suitable Grade Level

Advanced Undergraduate