/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Exercises \(23-26\) concern a ve... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Exercises \(23-26\) concern a vector space \(V,\) a basis \(\mathcal{B}=\) \(\left\\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{n}\right\\},\) and the coordinate mapping \(\mathbf{x} \mapsto[\mathbf{x}]_{\mathcal{B}}\) Show that a subset \(\left\\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\right\\} \quad\) in \(V\) is linearly independent if and only if the set of coordinate vectors \(\left\\{\left[\mathbf{u}_{1}\right]_{\mathcal{B}}, \ldots,\left[\mathbf{u}_{p}\right]_{\mathcal{B}}\right\\}\) is linearly independent in \(\mathbb{R}^{n} .[\text { Hint: }\) since the coordinate mapping is one-to-one, the following equations have the same solutions, \(c_{1}, \ldots, c_{p} . ]\)

Short Answer

Expert verified
The subset \(\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\}\) is linearly independent in \(V\) if and only if their coordinate vectors \(\{[\mathbf{u}_{1}]_{\mathcal{B}}, \ldots, [\mathbf{u}_{p}]_{\mathcal{B}}\}\) are linearly independent in \(\mathbb{R}^{n}\).

Step by step solution

01

Understand Coordinate Mapping

The coordinate mapping \([x]_{\mathcal{B}}\) assigns a vector \(\mathbf{x}\) in vector space \(V\) to a unique coordinate vector in \(\mathbb{R}^{n}\). This establishes a one-to-one correspondence between the vectors in \(V\) and their coordinate vectors under the basis \(\mathcal{B}\).
02

Define Linear Independence in V

A set of vectors \(\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\}\) in \(V\) is linearly independent if the only solution to the equation \(c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + \ldots + c_p \mathbf{u}_p = \mathbf{0}\) is the trivial solution, where all scalars \(c_1, c_2, \ldots, c_p\) are zero.
03

Apply Coordinate Mapping

Apply the coordinate mapping to the vector equation from Step 2 to obtain the coordinate vector equation: \(c_1 [\mathbf{u}_1]_{\mathcal{B}} + c_2 [\mathbf{u}_2]_{\mathcal{B}} + \ldots + c_p [\mathbf{u}_p]_{\mathcal{B}} = \mathbf{0}\) in \(\mathbb{R}^{n}\).
04

Linear Independence in R^n

A set of coordinate vectors \(\{[\mathbf{u}_{1}]_{\mathcal{B}}, \ldots, [\mathbf{u}_{p}]_{\mathcal{B}}\}\) is linearly independent in \(\mathbb{R}^{n}\) if the only solution to the equation in Step 3 is the trivial solution, meaning all coefficients must be zero.
05

Conclude Equivalence

Since the coordinate mapping is one-to-one, the set of vectors \(\{\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\}\) is linearly independent in \(V\) if and only if their corresponding coordinate vectors \(\{[\mathbf{u}_{1}]_{\mathcal{B}}, \ldots, [\mathbf{u}_{p}]_{\mathcal{B}}\}\) are linearly independent in \(\mathbb{R}^{n}\). Thus, linear independence is preserved under the coordinate mapping.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Coordinate Mapping
In the realm of vector spaces, coordinate mapping is a crucial concept. Essentially, it is a method of translating vectors from an abstract vector space, denoted as \( V \), to a more concrete form in \( \mathbb{R}^{n} \). This is done using a specified basis, \( \mathcal{B} = \{ \mathbf{b}_1, \ldots, \mathbf{b}_n \} \).

The term "coordinate vector" or "coordinates of a vector" refers to expressing any vector \( \mathbf{x} \) in terms of its coordinates relative to the basis \( \mathcal{B} \). The coordinate mapping \( \mathbf{x} \mapsto [\mathbf{x}]_{\mathcal{B}} \) captures this idea by mapping each vector \( \mathbf{x} \) in \( V \) to a unique coordinate vector \([\mathbf{x}]_{\mathcal{B}} \) in \( \mathbb{R}^{n} \). This unique correspondence is one-to-one, meaning each vector in \( V \) has a single, corresponding coordinate vector.

Coordinate mapping is like translating a language; it converts the vectors into a numerical sequence that retains all original properties, allowing easier computation and analysis within \( \mathbb{R}^{n} \).
Basis of a Vector Space
The concept of a basis is pivotal to understanding vector spaces. A basis \( \mathcal{B} = \{ \mathbf{b}_1, \ldots, \mathbf{b}_n \} \) for a vector space \( V \) is a set of vectors that are both linearly independent and span the entire space. This means that any vector in the space can be written uniquely as a linear combination of the basis vectors.

To illustrate, consider a vector space \( V \) which could be imagined as all the possible directions and magnitudes you might travel from a point. A basis within this space acts like a set of axes in a coordinate system. Each vector in \( V \) can be described by how far it stretches along each axis (i.e., how much of each basis vector it contains).

Determining a basis is crucial because it simplifies many problems. Once a basis is chosen, any problem within the vector space can be tackled within the familiar confines of \( \mathbb{R}^{n} \), using the compact notation and powerful tools of coordinate vectors.
Linear Independence in \( \mathbb{R}^{n} \)
Linear independence is a fundamental property of a set of vectors, whether in an abstract space \( V \) or in \( \mathbb{R}^{n} \). A set of vectors \( \{ \mathbf{v}_1, \ldots, \mathbf{v}_p \} \) in any vector space is said to be linearly independent if the only solution to the equation:
  • \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \ldots + c_p \mathbf{v}_p = \mathbf{0} \)
is when all coefficients \( c_1, c_2, \ldots, c_p \) are zero.

In the context of \( \mathbb{R}^{n} \) specifically, linear independence involves sequences of numbers rather than abstract vectors. If a set of coordinate vectors in \( \mathbb{R}^{n} \) is linearly independent, it implies no vector in the set can be expressed as a combination of others. This property is essential for defining a basis in this space.

Identifying linearly independent vectors is crucial because it ensures that the coordinate vectors span the space effectively without redundancy. This is integral to problems like the one addressed here, where transforming the problem from an abstract vector space \( V \) to \( \mathbb{R}^{n} \) hinges upon verifying the linear independence of coordinate vectors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

[M] Examine powers of a regular stochastic matrix. a. Compute \(P^{k}\) for \(k=2,3,4,5,\) when $$ P=\left[\begin{array}{cccc}{.3355} & {.3682} & {.3067} & {.0389} \\ {.2663} & {.2723} & {.3277} & {.5451} \\ {.1935} & {.1502} & {.1589} & {.2395} \\\ {.2047} & {.2093} & {.2067} & {.1765}\end{array}\right] $$ Display calculations to four decimal places. What happens to the columns of \(P^{k}\) as \(k\) increases? Compute the steady-state vector for \(P\) . b. Compute \(Q^{k}\) for \(k=10,20, \ldots, 80,\) when $$ Q=\left[\begin{array}{ccc}{.97} & {.05} & {.10} \\ {0} & {.90} & {.05} \\\ {.03} & {.05} & {.85}\end{array}\right] $$ (Stability for \(Q^{k}\) to four decimal places may require \(k=116\) or more.) Compute the steady-state vector for \(Q .\) Conjecture what might be true for any regular stochastic matrix. c. Use Theorem 18 to explain what you found in parts (a) and \((b) .\)

If the null space of a \(5 \times 6\) matrix \(A\) is 4 -dimensional, what is the dimension of the column space of \(A ?\)

In Exercises 21 and \(22,\) mark each statement True or False. Justify each answer. a. A single vector by itself is linearly dependent. b. If \(H=\operatorname{Span}\left\\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{p}\right\\},\) then \(\left\\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{p}\right\\}\) is a basis for Ine columns of an invertible \(n \times n\) matrix form a basis c. The columns of an invertible \(n \times n\) matrix form a basis for \(\mathbb{R}^{n}\) . d. A basis is a spanning set that is as large as possible. e. In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix.

Exercises 23 and 24 refer to a difference equation of the form \(y_{k+1}-a y_{k}=b,\) for suitable constants \(a\) and \(b\) A loan of \(\$ 10,000\) has an interest rate of 1\(\%\) per month and a monthly payment of \(\$ 450 .\) The loan is made at month \(k=0\) , and the first payment is made one month later, at \(k=1 .\) For \(k=0,1,2, \ldots,\) let \(y_{k}\) be the unpaid balance of the loan just after the \(k\) th monthly payment. Thus a. Write a difference equation satisfied by \(\left\\{y_{k}\right\\}\) b. \([\mathbf{M}]\) Create a table showing \(k\) and the balance \(y_{k}\) at month \(k .\) List the program or the keystrokes you used to create the table. c. \([\mathbf{M}]\) What will \(k\) be when the last payment is made? How much will the last payment be? How much money did the borrower pay in total?

In Exercises \(25-28\) , show that the given signal is a solution of the difference equation. Then find the general solution of that difference equation. $$ y_{k}=2 k-4 ; y_{k+2}+\frac{3}{2} y_{k+1}-y_{k}=1+3 k $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.