Chapter 3: Problem 5
Let \(S\) be the collection of vectors \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]\) in \(\mathbb{R}^{3}\) that satisfy the given property. In each case, either prove that S forms a subspace of \(\mathbb{R}^{3}\) or give a counterexample to show that it does not. $$x=y=z$$
Short Answer
Expert verified
Yes, \( S \) forms a subspace of \( \mathbb{R}^3 \).
Step by step solution
01
Understand the Problem
We start by analyzing the condition given for the vectors. The problem states that each vector in the set \( S \) satisfies \( x = y = z \). This means, for any vector \( \mathbf{v} \in S \), it has the form \( \mathbf{v} = \begin{bmatrix} x \ x \ x \end{bmatrix} \). Now, we need to determine if these vectors form a subspace of \( \mathbb{R}^3 \).
02
Check for Zero Vector
A subspace must contain the zero vector. The zero vector in \( \mathbb{R}^3 \) is \( \mathbf{0} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \). Substituting \( x=0 \) into the form of the vector, we get \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \), which is clearly in \( S \). Thus, \( S \) contains the zero vector.
03
Check for Closure under Addition
To be a subspace, the set \( S \) must be closed under vector addition. Take two vectors \( \mathbf{u} = \begin{bmatrix} x \ x \ x \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} y \ y \ y \end{bmatrix} \) in \( S \). Their sum is \( \mathbf{u} + \mathbf{v} = \begin{bmatrix} x+y \ x+y \ x+y \end{bmatrix} \). Since \( x+y = x+y = x+y \), the sum is also of the form \( \begin{bmatrix} z \ z \ z \end{bmatrix} \) for \( z=x+y \). Thus, \( \mathbf{u} + \mathbf{v} \in S \), verifying closure under addition.
04
Check for Closure under Scalar Multiplication
The set \( S \) also needs to be closed under scalar multiplication. Consider a vector \( \mathbf{u} = \begin{bmatrix} x \ x \ x \end{bmatrix} \in S \) and a scalar \( c \). The scalar multiplication \( c\mathbf{u} = c \begin{bmatrix} x \ x \ x \end{bmatrix} = \begin{bmatrix} cx \ cx \ cx \end{bmatrix} \). Since \( cx = cx = cx \), \( c\mathbf{u} \) remains in the form \( \begin{bmatrix} z \ z \ z \end{bmatrix} \), so \( c\mathbf{u} \in S \). This verifies closure under scalar multiplication.
05
Conclusion on Subspace Status
Since \( S \) contains the zero vector, is closed under addition, and closed under scalar multiplication, \( S \) satisfies all the conditions to be a subspace of \( \mathbb{R}^3 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Addition
Vector addition is a fundamental operation in vector spaces, including subspaces. It involves adding two vectors together to produce a third vector. Imagine you have two vectors, \( \mathbf{a} \) and \( \mathbf{b} \), both belonging to a certain set \( S \). For \( S \) to be considered a subspace, their sum \( \mathbf{a} + \mathbf{b} \) must also be in \( S \).
In our case, the set \( S \) consists of vectors of the form \( \begin{bmatrix} x \ x \ x \end{bmatrix} \). This means that for any two vectors in \( S \), say \( \mathbf{u} = \begin{bmatrix} x \ x \ x \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} y \ y \ y \end{bmatrix} \), their sum is \( \mathbf{u} + \mathbf{v} = \begin{bmatrix} x+y \ x+y \ x+y \end{bmatrix} \). This resulting vector is still of the form \( \begin{bmatrix} z \ z \ z \end{bmatrix} \), showing that it remains in \( S \).
In our case, the set \( S \) consists of vectors of the form \( \begin{bmatrix} x \ x \ x \end{bmatrix} \). This means that for any two vectors in \( S \), say \( \mathbf{u} = \begin{bmatrix} x \ x \ x \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} y \ y \ y \end{bmatrix} \), their sum is \( \mathbf{u} + \mathbf{v} = \begin{bmatrix} x+y \ x+y \ x+y \end{bmatrix} \). This resulting vector is still of the form \( \begin{bmatrix} z \ z \ z \end{bmatrix} \), showing that it remains in \( S \).
- This confirms the subspace property of being closed under addition.
- Vector addition ensures that the structure of a subspace is maintained when vectors within it are combined.
Scalar Multiplication
Scalar multiplication involves multiplying each component of a vector by a scalar (a real number). It is another vital operation for any subspace within a vector space. For a set to qualify as a subspace, it must be closed under scalar multiplication.
When focusing on our example with set \( S \), each vector is of the form \( \begin{bmatrix} x \ x \ x \end{bmatrix} \). When you multiply this vector by a scalar \( c \), the result is \( c \begin{bmatrix} x \ x \ x \end{bmatrix} = \begin{bmatrix} cx \ cx \ cx \end{bmatrix} \). Since this new form \( \begin{bmatrix} z \ z \ z \end{bmatrix} \) remains within \( S \), closure under scalar multiplication is verified.
When focusing on our example with set \( S \), each vector is of the form \( \begin{bmatrix} x \ x \ x \end{bmatrix} \). When you multiply this vector by a scalar \( c \), the result is \( c \begin{bmatrix} x \ x \ x \end{bmatrix} = \begin{bmatrix} cx \ cx \ cx \end{bmatrix} \). Since this new form \( \begin{bmatrix} z \ z \ z \end{bmatrix} \) remains within \( S \), closure under scalar multiplication is verified.
- This property shows that every scalar multiple of a vector in the subspace is still part of the subspace.
- Scalar multiplication helps to scale vectors up or down while retaining their alignment within the subspace.
Zero Vector
The zero vector is a unique vector where all components are zero, such as \( \mathbf{0} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \) in \( \mathbb{R}^3 \). A subspace must contain the zero vector to adhere to its mathematical definition.
In our example, ensuring the zero vector is part of set \( S \) involves substituting zero for all components in the vectors of the form \( \begin{bmatrix} x \ x \ x \end{bmatrix} \). Doing so yields \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \), which is precisely the zero vector required in \( S \).
In our example, ensuring the zero vector is part of set \( S \) involves substituting zero for all components in the vectors of the form \( \begin{bmatrix} x \ x \ x \end{bmatrix} \). Doing so yields \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \), which is precisely the zero vector required in \( S \).
- Inclusion of the zero vector is crucial because it represents the additive identity in vector spaces.
- It establishes the baseline for other vectors, ensuring that they can return to a starting, neutral point through operations like addition.