Question

We often write vectors in \(\mathbb{R}^{n}\) as rows. In each case either show that the statement is true or give an example showing that it is false. a. If \(U \neq \mathbb{R}^{n}\) is a subspace of \(\mathbb{R}^{n}\) and \(\mathbf{x}+\mathbf{y}\) is in \(U\) then \(\mathbf{x}\) and \(\mathbf{y}\) are both in \(U\). b. If \(U\) is a subspace of \(\mathbb{R}^{n}\) and \(r \mathbf{x}\) is in \(U\) for all \(r\) in \(\mathbb{R},\) then \(\mathbf{x}\) is in \(U\) c. If \(U\) is a subspace of \(\mathbb{R}^{n}\) and \(\mathbf{x}\) is in \(U,\) then \(-\mathbf{x}\) is also in \(U\) d. If \(\mathbf{x}\) is in \(U\) and \(U=\operatorname{span}\\{\mathbf{y}, \mathbf{z}\\},\) then \(U=\) \(\operatorname{span}\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\)

Short answer

a. False; b. True; c. True; d. True.

Step-by-step solution

Analyzing Statement a

The statement says that if a subspace \( U \) of \( \mathbb{R}^{n} \) does not equal \( \mathbb{R}^{n} \) and the sum \( \mathbf{x} + \mathbf{y} \) belongs to \( U \), then \( \mathbf{x} \) and \( \mathbf{y} \) must both be in \( U \). This statement is false. To disprove it, consider a specific example: let \( U = \{ (0, 0) \} \subset \mathbb{R}^2 \). Clearly, \( U eq \mathbb{R}^{2} \). The zero vector \( \mathbf{0} = \mathbf{x} + \mathbf{y} \) is in \( U \), but neither \( \mathbf{x} = (1, 0) \) nor \( \mathbf{y} = (-1, 0) \) is in \( U \).

Analyzing Statement b

The statement claims that if \( r\mathbf{x} \) belongs to a subspace \( U \) for all real numbers \( r \), then \( \mathbf{x} \) must be in \( U \). This statement is true. If \( r\mathbf{x} \) is in \( U \) for all \( r \), we can take \( r = 1 \) to find that \( \mathbf{x} = 1\mathbf{x} \) is in \( U \). Therefore, \( \mathbf{x} \) must be in \( U \).

Analyzing Statement c

The statement asserts that if \( \mathbf{x} \) is in a subspace \( U \) of \( \mathbb{R}^{n} \), then the negation \( -\mathbf{x} \) is also in \( U \). This statement is true. One of the properties of a subspace is that it is closed under scalar multiplication. Since \( -1 \) is a scalar, \( -\mathbf{x} = (-1)\mathbf{x} \) is also in \( U \) if \( \mathbf{x} \) is in \( U \).

Analyzing Statement d

The statement suggests that if \( \mathbf{x} \) is in subspace \( U \) and \( U = \operatorname{span}\{\mathbf{y}, \mathbf{z}\} \), then \( U = \operatorname{span}\{\mathbf{x}, \mathbf{y}, \mathbf{z}\} \). This statement is true. Since \( \mathbf{x} \) lies in \( U = \operatorname{span}\{\mathbf{y}, \mathbf{z}\} \), \( \mathbf{x} \) can be expressed as a linear combination of \( \mathbf{y} \) and \( \mathbf{z} \). So adding \( \mathbf{x} \) to the spanning set \( \{\mathbf{y}, \mathbf{z}\} \) does not increase the dimension of the span, hence \( U \) remains \( \operatorname{span}\{\mathbf{y}, \mathbf{z}\} \).

Key concepts

Vector Spaces

In linear algebra, a vector space is a set of vectors that can be added together and multiplied by scalars, following certain rules. A vector in the vector space must be able to meet these two operations, vector addition and scalar multiplication, without leaving the space. This means for any vectors \( \mathbf{a} \) and \( \mathbf{b} \) in the vector space, their sum \( \mathbf{a} + \mathbf{b} \) should also be in the vector space. Furthermore, any vector multiplied by a scalar from the underlying field (like real numbers \( \mathbb{R} \)) must still be a vector in that space. This characteristic is known as closure.
Importance of vector spaces cannot be overstated, as they serve as a fundamental concept in mathematics, particularly in linear algebra.

• They offer a structured way to work with complex systems mathematically.
• They are utilized in various applications like computer graphics, engineering, and physics.

Understanding vector spaces helps in grasping how different dimensions relate within different mathematical systems.

Scalar Multiplication

Scalar multiplication is the process of multiplying a vector by a scalar. A scalar is a single number from the set of real numbers \( \mathbb{R} \) in many instances. When we multiply a vector by a scalar, every component of the vector is multiplied by that scalar. For example, if you have a vector \( \mathbf{v} = (v_1, v_2, v_3) \) and a scalar \( c \), the result of the scalar multiplication is \( c \cdot \mathbf{v} = (c \cdot v_1, c \cdot v_2, c \cdot v_3) \).
Before applying scalar multiplication, it's important to recognize that the resulting vector should remain within the same vector space, preserving its structural integrity. This closure property ensures that the vector space remains intact regardless of the scalar used.

• Scalar multiplication affects the length and direction of the vectors but does not change the fundamental nature of the vector space.
• This operation is crucial in defining operations in vector spaces such as span and linear transformations.

Linear Combinations

A linear combination involves creating a new vector by adding together several vectors that have been scaled by different scalars. If you have a set of vectors \( \{\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_n \} \) and corresponding scalars \( c_1, c_2, \ldots, c_n \), a linear combination would be expressed as \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \ldots + c_n \mathbf{v}_n \).
Linear combinations are central to discussing concepts like the span of a collection of vectors. The span is essentially all possible linear combinations of a set of vectors. This concept helps define subspaces, as a subspace can be described as a span of vectors.

• Linear combinations allow us to express vectors flexibly, combining multiple influences.
• They are fundamental in solving systems of linear equations and in defining vector spaces and subspaces.

Understanding linear combinations enable us to appreciate how vectors interrelate and interact within a vector space.