Chapter 4: Problem 11
Let \(W\) be the set of all vectors of the form \(\left[\begin{array}{c}{5 b+2 c} \\\ {b} \\ {c}\end{array}\right]\), where \(b\) and \(c\) are arbitrary. Find vectors \(\mathbf{u}\) and \(\mathbf{v}\) such that \(W=\operatorname{Span}\\{\mathbf{u}, \mathbf{v}\\} .\) Why does this show that \(W\) is a subspace of \(\mathbb{R}^{3} ?\)
Short Answer
Step by step solution
Understand the Problem
Express the Given Vector in Terms of Basis Components
Identify the Vectors to Span W
Verify Subspace Properties
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Spaces
A vector space must satisfy several properties for these operations:
- **Commutative Law for Addition**: For any two vectors \( \mathbf{u} \) and \( \mathbf{v} \) in the space, \( \mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u} \).
- **Associative Law for Addition**: For any vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \), \( (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) \).
- **Existence of Zero Vector**: There is a vector, often written as \( \mathbf{0} \), such that \( \mathbf{u} + \mathbf{0} = \mathbf{u} \) for any vector \( \mathbf{u} \).
- **Existence of Additive Inverses**: For every vector \( \mathbf{u} \), there is a vector \( -\mathbf{u} \) such that \( \mathbf{u} + (-\mathbf{u}) = \mathbf{0} \).
- **Distributive Laws for Scalar Multiplication**: \( a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v} \) and \( (a + b)\mathbf{u} = a\mathbf{u} + b\mathbf{u} \) for scalars \(a\) and \(b\).
- **Compatibility of Scalar Multiplication**: \( a(b\mathbf{u}) = (ab)\mathbf{u} \).
- **Identity of Scalar Multiplication**: \( 1\mathbf{u} = \mathbf{u} \) for any vector \( \mathbf{u} \).
Subspaces
- **Contains the Zero Vector**: The zero vector must be part of the subspace. This ensures non-emptiness.
- **Closure under Addition**: If you take any two vectors from the subspace and add them, the result must also be a vector within the same subspace.
- **Closure under Scalar Multiplication**: Multiplying any vector in the subspace by a scalar produces another vector that remains in the subspace.
Linear Combinations
Given the vectors:
- \( \mathbf{u} = \begin{bmatrix} 5 \ 1 \ 0 \end{bmatrix} \)
- \( \mathbf{v} = \begin{bmatrix} 2 \ 0 \ 1 \end{bmatrix} \)
This representation shows how every vector in \( W \) can be built using these selected vectors and appropriate scalar multipliers.
Basis and Span
For the set \( W = \operatorname{Span}\{ \mathbf{u}, \mathbf{v} \} \), where \( \mathbf{u} = \begin{bmatrix} 5 \ 1 \ 0 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} 2 \ 0 \ 1 \end{bmatrix} \):
- **Span**: Consists of all vectors that you can generate using linear combinations of \( \mathbf{u} \) and \( \mathbf{v} \).
- **Basis**: A set of vectors that not only spans a space but is also linearly independent. This means no vector in the set can be written as a linear combination of the others.