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Chapter 5: Problem 2

We write vectors \(\mathbb{R}^{n}\) as rows. Let \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\\}\) be an independent set in \(\mathbb{R}^{n} .\) Which of the following sets is independent? Support your answer. a. \(\\{\mathbf{x}-\mathbf{y}, \mathbf{y}-\mathbf{z}, \mathbf{z}-\mathbf{X}\\}\) b. \(\\{\mathbf{x}+\mathbf{y}, \mathbf{y}+\mathbf{z}, \mathbf{z}+\mathbf{x}\\}\) c. \(\\{\mathbf{x}-\mathbf{y}, \mathbf{y}-\mathbf{z}, \mathbf{z}-\mathbf{w}, \mathbf{w}-\mathbf{x}\\}\) d. \(\\{\mathbf{x}+\mathbf{y}, \mathbf{y}+\mathbf{z}, \mathbf{z}+\mathbf{w}, \mathbf{w}+\mathbf{x}\\}\)

Short Answer

Expert verified
Sets a, b, and c are independent. Set d is dependent.

Step by step solution

01

Understanding Independence

A set of vectors is independent if no vector can be written as a linear combination of the others. Since \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\}\) is independent, no vector is a linear combination of the others.
02

Analyzing Set a

Set a: \(\{\mathbf{x}-\mathbf{y}, \mathbf{y}-\mathbf{z}, \mathbf{z}-\mathbf{x}\}\). The set has three vectors formed by differences. Suppose \(a(\mathbf{x}-\mathbf{y}) + b(\mathbf{y}-\mathbf{z}) + c(\mathbf{z}-\mathbf{x}) = \mathbf{0}\). Simplify this to get: \((a-c)\mathbf{x} + (b-a)\mathbf{y} + (c-b)\mathbf{z} = \mathbf{0}\). For them to be 0, \(a = b = c\). Therefore, a is independent.
03

Analyzing Set b

Set b: \(\{\mathbf{x}+\mathbf{y}, \mathbf{y}+\mathbf{z}, \mathbf{z}+\mathbf{x}\}\). Consider \(a(\mathbf{x}+\mathbf{y}) + b(\mathbf{y}+\mathbf{z}) + c(\mathbf{z}+\mathbf{x}) = \mathbf{0}\). Simplify: \((a+c)\mathbf{x} + (a+b)\mathbf{y} + (b+c)\mathbf{z} = \mathbf{0}\). This results in equations \(a+c=0\), \(a+b=0\), \(b+c=0\), leading to \(a=b=c=0\). Thus, b is independent.
04

Analyzing Set c

Set c: \(\{\mathbf{x}-\mathbf{y}, \mathbf{y}-\mathbf{z}, \mathbf{z}-\mathbf{w}, \mathbf{w}-\mathbf{x}\}\). Suppose \(a(\mathbf{x}-\mathbf{y}) + b(\mathbf{y}-\mathbf{z}) + c(\mathbf{z}-\mathbf{w}) + d(\mathbf{w}-\mathbf{x}) = \mathbf{0}\). Simplify: \((a-d)\mathbf{x} + (b-a)\mathbf{y} + (c-b)\mathbf{z} + (d-c)\mathbf{w} = \mathbf{0}\). We get \(a = b = c = d = 0\), indicating independence.
05

Analyzing Set d

Set d: \(\{\mathbf{x}+\mathbf{y}, \mathbf{y}+\mathbf{z}, \mathbf{z}+\mathbf{w}, \mathbf{w}+\mathbf{x}\}\). Consider \(a(\mathbf{x}+\mathbf{y}) + b(\mathbf{y}+\mathbf{z}) + c(\mathbf{z}+\mathbf{w}) + d(\mathbf{w}+\mathbf{x}) = \mathbf{0}\). Simplify: \((a+d)\mathbf{x} + (a+b)\mathbf{y} + (b+c)\mathbf{z} + (c+d)\mathbf{w} = \mathbf{0}\). These conditions do not uniquely determine \(a, b, c, d = 0\), indicating dependence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Combinations
In linear algebra, a linear combination refers to an expression made up of sums and scalar multiplications involving vectors. Whenever we discuss vectors, it's crucial to understand how these combinations work. For instance, if you have vectors \(\mathbf{x}, \mathbf{y},\) and \(\mathbf{z}\), a linear combination of these vectors can be written as \(a\mathbf{x} + b\mathbf{y} + c\mathbf{z}\), where \(a, b,\) and \(c\) are scalars.
One of the key points is when a set of vectors is independent. No vector in this set can be expressed as a linear combination of the others. This condition holds true for the independent set \(\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{w}\}\) discussed in the exercise. Similarly, in the step-by-step solution, sets that lead to expressions equating to \(\mathbf{0}\) without forcing any scalar other than zero, show independence.
Understanding and forming sets of independent vectors through linear combinations are fundamental concepts in linear algebra and have numerous applications in mathematics and science.
Linear Algebra
Linear algebra encompasses various mathematical concepts including vectors, matrices, and vector spaces. It's one of the foundational branches of mathematics used to deal with linear equations and their representations via vectors and matrices.
Central to linear algebra is the idea of vector independence. As seen in the exercise, the independence of vectors ensures that solutions to equations are unique. For example, if a group of vectors is independent, it means no vector in the group can be constructed by linearly combining any of the others. This is akin to having unique solutions in systems of linear equations.
Linear algebra not only facilitates the understanding of vector spaces and transformations but also extends to solving practical problems in engineering, physics, computer science, and more. Recognizing dependent versus independent vectors is critical when working with linear transformations and vector spaces.
Vector Spaces
Vector spaces form the backbone of linear algebra and are crucial for understanding the behavior of vectors in any mathematical context. A vector space consists of vectors and is equipped with operations of vector addition and scalar multiplication that satisfy eight axioms, such as associativity, commutativity, and distributivity.
In any vector space, the concept of independence is significant. Independent vectors are those that satisfy none of their members being a linear combination of others. This independence indicates a maximum set of vectors, known as the basis, which fully spans the vector space.
In the exercise, the challenge was to determine the independence of various vector sets. Understanding whether a collection of vectors within a space is independent or dependent determines how that space is structured and paved the way for practical applications. Vector spaces manifest in numerous real-world applications where data and functions are interpreted as vectors.

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Most popular questions from this chapter

Let \(\lambda\) be an eigenvalue of \(A\) with corresponding eigenvector \(\mathbf{x}\). If \(B=P^{-1} A P\) is similar to \(A\), show that \(P^{-1} \mathbf{x}\) is an eigenvector of \(B\) corresponding to \(\lambda\)

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(A\) is an \(m \times n\) matrix, show that, for each invertible \(n \times n\) matrix \(V, \operatorname{im}(A)=\operatorname{im}(A V)\)

Use the Cauchy inequality to show that \(\sqrt{x y} \leq \frac{1}{2}(x+y)\) for all \(x \geq 0\) and \(y \geq 0\). Here \(\sqrt{x y}\) and \(\frac{1}{2}(x+y)\) are called, respectively, the geometric mean and arithmetic mean of \(x\) and \(y .\) [Hint: Use \(\mathbf{x}=\left[\begin{array}{l}\sqrt{x} \\\ \sqrt{y}\end{array}\right]\) and \(\left.\mathbf{y}=\left[\begin{array}{c}\sqrt{y} \\\ \sqrt{x}\end{array}\right] .\right]\)

Let \(P\) be an invertible \(n \times n\) matrix. If \(A\) is any \(n \times n\) matrix, write \(T_{P}(A)=P^{-1} A P\). Verify that: a. \(T_{P}(I)=I\) b. \(T_{P}(A B)=T_{P}(A) T_{P}(B)\) c. \(T_{P}(A+B)=T_{P}(A)+T_{P}(B)\) d. \(T_{P}(r A)=r T_{P}(A)\) e. \(T_{P}\left(A^{k}\right)=\left[T_{P}(A)\right]^{k}\) for \(k \geq 1\) f. If \(A\) is invertible, \(T_{P}\left(A^{-1}\right)=\left[T_{P}(A)\right]^{-1}\). g. If \(Q\) is invertible, \(T_{Q}\left[T_{P}(A)\right]=T_{P Q}(A)\).

Show that \(\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\ 2 & 0 & 1 & 1 \\ 1 & 1 & 0 & -1 \\ 4 & 3 & 0 & 0\end{array}\right]\) and \(\left[\begin{array}{rrrr}1 & -1 & 3 & 0 \\ -1 & 0 & 1 & 1 \\ 0 & -1 & 4 & 1 \\\ 5 & -1 & -1 & -4\end{array}\right]\) are not similar.
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