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Chapter 2: Problem 48

Let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) be a linearly independent set of vectors in \(\mathbb{R}^{n}\), andlet v be a vector in \(\mathbb{R}^{n}\). Suppose that \(\mathbf{v}=c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{k} \mathbf{v}_{k}\) with \(c_{1} \neq 0 .\) Prove that \(\left\\{\mathbf{v}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\right\\}\) is linearly independent

Short Answer

Expert verified
The set \( \{\mathbf{v}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\} \) is linearly independent because the only solution is \( a = b_2 = \cdots = b_k = 0 \).

Step by step solution

01

Understand Linear Independence

A set of vectors \( \{\mathbf{u}_1, \mathbf{u}_2, \ldots, \mathbf{u}_m\} \) is linearly independent if the only solution to the equation \( a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2 + \cdots + a_m \mathbf{u}_m = \mathbf{0} \) is \( a_1 = a_2 = \cdots = a_m = 0 \).
02

Examine the Given Vector Expression

We are given that \( \mathbf{v} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_k \mathbf{v}_k \) where \( c_1 eq 0 \). This expression shows that \( \mathbf{v} \) is a linear combination of the vectors \( \{\mathbf{v}_1, \ldots, \mathbf{v}_k\} \).
03

Consider the Set Including \( \mathbf{v} \)

We need to prove that the set \( \{\mathbf{v}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\} \) is linearly independent. Assume the linear combination: \( a\mathbf{v} + b_2\mathbf{v}_2 + \cdots + b_k\mathbf{v}_k = \mathbf{0} \).
04

Substitute \( \mathbf{v} \) in the Equation

Substitute the expression for \( \mathbf{v} \) into the linear combination, giving: \( a(c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_k \mathbf{v}_k) + b_2\mathbf{v}_2 + \cdots + b_k\mathbf{v}_k = \mathbf{0} \).
05

Simplify the Combined Equation

By distributing \( a \), the equation becomes: \( ac_1 \mathbf{v}_1 + (ac_2 + b_2) \mathbf{v}_2 + \cdots + (ac_k + b_k) \mathbf{v}_k = \mathbf{0} \).
06

Analyze the Coefficients for Zero Solution

Since the original set \( \{\mathbf{v}_1, \ldots, \mathbf{v}_k\} \) is linearly independent, the coefficients of each vector must be zero: \( ac_1 = 0 \), \( ac_2 + b_2 = 0 \), ... , \( ac_k + b_k = 0 \).
07

Solve for Each Coefficient

Since \( c_1 eq 0 \), \( ac_1 = 0 \) implies \( a = 0 \). Substituting \( a = 0 \) into the other equations, we have \( b_2 = b_3 = \cdots = b_k = 0 \).
08

Conclude Linearly Independent Set

Since \( a = b_2 = \cdots = b_k = 0 \) is the only solution, \( \{\mathbf{v}, \mathbf{v}_{2}, \ldots, \mathbf{v}_{k}\} \) is linearly independent.

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Key Concepts

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

Vector Spaces
Vector spaces are fundamental concepts in linear algebra, serving as the environment in which vectors reside and operate. These structures consist of a set of vectors, which can be added together and multiplied by scalars, provided they satisfy specific rules and properties.
A vector space must follow a few essential criteria:
  • Vector Addition: For any two vectors in the space, their sum must also be a vector in the space.
  • Scalar Multiplication: Multiplying any vector in the space by a scalar should yield another vector within the same space.
  • Zero Vector: The space must include a zero vector (a vector with all components equal to zero).
  • Distributive and Associative Properties: These ensure logical behaviors for vector addition and multiplication.
Understanding vector spaces is crucial because they provide a framework to study vectors systematically. They form the basis for dealing with more complex structures like linear transformations and matrix operations.
Linear Combinations
Linear combinations play a vital role when working with vectors. The term refers to the expression created by multiplying vectors by scalars and then adding the results together.
Keeping this concept in mind, if you are given vectors \( \mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k \), any vector \( \mathbf{v} \) that can be expressed in the form:\[ \mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k \] is derived through a linear combination. Here, \( c_1, c_2, \ldots, c_k \) are scalar coefficients.
Linear combinations are essential because they provide a mechanism to generate new vectors from existing ones. This helps in understanding vector spaces, dimensions, and examining relationships such as linear independence or dependence.
R^n
The notation \( \mathbb{R}^n \) refers to the n-dimensional real coordinate space. It is an example of a vector space where each of the vectors has "n" components, each of which is a real number.
This space is crucial in linear algebra because:
  • It allows the analysis of systems of equations and linear inequalities.
  • Any point in this space corresponds to a coordinate vector with n real numbers.
  • It forms the setting for geometric interpretations of linear transformations.
For instance, in \( \mathbb{R}^3 \), vectors are used to represent points or directions in three-dimensional space, allowing for practical applications like physics simulations and 3D graphics.
Overall, \( \mathbb{R}^n \) forms a fundamental framework for understanding more advanced topics in mathematics and applied sciences.

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Most popular questions from this chapter

What is wrong with the following "proof" that every matrix with at least two rows is row equivalent to a matrix with a zero row? Perform \(R_{2}+R_{1}\) and \(R_{1}+R_{2}\). Now rows 1 and 2 are identical. Now perform \(R_{2}-R_{1}\) to obtain a row of zeros in the second row.

Determine whether the lines \(\mathbf{x}=\mathbf{p}+\) su and \(\mathbf{x}=\mathbf{q}+t \mathbf{v}\) intersect and, if they do, find their point of intersection. \(\mathbf{p}=\left[\begin{array}{l}3 \\ 1 \\ 0\end{array}\right], \mathbf{q}=\left[\begin{array}{r}-1 \\ 1 \\ -1\end{array}\right], \mathbf{u}=\left[\begin{array}{l}1 \\ 0 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{l}2 \\ 3 \\ 1\end{array}\right]\)

Compute the first four iterates, using the zero vector as the initial approximation, to show that the Gauss-Seidel method diverges. Then show that the equations can be rearranged to give a strictly diagonally dominant coefficient matrix, and apply the Gauss-Seidel method to obtain an approximate solution that is accurate to within 0.001. $$\begin{aligned}x_{1}-4 x_{2}+2 x_{3} &=2 \\\2 x_{2}+4 x_{3} &=1 \\\6 x_{1}-x_{2}-2 x_{3} &=1\end{aligned}$$

In Gotham City, the departments of Administration (A), Health (H), and Transportation (T) are interdependent. For every dollar's worth of services they produce, each department uses a certain amount of the services produced by the other departments and itself, as shown in Table \(2.10 .\) Suppose that, during the year, other city departments require \(\$ 1\) million in Administrative services, \(\$ 1.2\) million in Health services, and \(\$ 0.8\) million in Transportation services. What does the annual dollar value of the services produced by each department need to be in order to meet the demands?

(a) If the columns of an \(n \times n\) matrix \(A\) are linearly independent as vectors in \(\mathbb{R}^{n}\), what is the rank of \(A\) ? Explain. (b) If the rows of an \(n \times n\) matrix \(A\) are linearly independent as vectors in \(\mathbb{R}^{n},\) what is the rank of \(A\) ? Explain.
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