/*! 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 3 Let \(H\) be the set of points i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(H\) be the set of points inside and on the unit circle in the \(x y\) -plane. That is, let \(H=\left\\{\left[\begin{array}{c}{x} \\\ {y}\end{array}\right] : x^{2}+y^{2} \leq 1\right\\} .\) Find a specific example - two vectors or a vector and a scalar - to show that \(H\) is not a subspace of \(\mathbb{R}^{2}\) .

Short Answer

Expert verified
\(H\) is not a subspace because it is not closed under scalar multiplication.

Step by step solution

01

Identify Conditions for Subspace

To show that a set is a subspace of \( \mathbb{R}^2 \), it must satisfy three conditions: it contains the zero vector, it is closed under vector addition, and it is closed under scalar multiplication. We need to check whether \( H \) satisfies these conditions.
02

Check Zero Vector Inclusion

The zero vector in \( \mathbb{R}^2 \) is \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Check if this vector lies in set \( H \). Since \( 0^2 + 0^2 = 0 \leq 1 \), \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) is in \( H \). Thus, \( H \) contains the zero vector.
03

Test for Closure Under Scalar Multiplication

Choose a point in \( H \), such as \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \). Check if a scalar multiple of this vector is in \( H \). At scalar 2, the vector becomes \( \begin{bmatrix} 2 \ 0 \end{bmatrix} \), and \( 2^2 + 0^2 = 4 \), which is greater than 1. Hence, \( H \) is not closed under scalar multiplication.
04

Conclusion

Since \( H \) is not closed under scalar multiplication, it cannot be a subspace of \( \mathbb{R}^2 \). Just verifying one failed condition is sufficient to establish this conclusion.

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.

Vector Spaces
A vector space is a mathematical structure formed by a collection of vectors, where two operations—vector addition and scalar multiplication—are defined and satisfy certain conditions. Vector spaces are foundational in linear algebra and can exist over any field of numbers, such as real or complex numbers. To understand whether a specific set can be considered a vector space, it is essential to check if it adheres to specific properties. These properties include:
  • The presence of an additive identity, typically the zero vector.
  • Closure under vector addition, where the sum of any two vectors in the set results in another vector within the same set.
  • Closure under scalar multiplication, meaning multiplying any vector in the set by a scalar yields another vector in the same set.
It is this structure and these properties that allow vector spaces to be versatile in applications across physics, computer science, and more.
Subspaces
Subspaces are crucial components within vector spaces. A subspace is simply a subset that is also a vector space in its own right. In our exploration of vector spaces, subspaces have to fulfill the same properties with a slight focus on the encompassing vector space. To identify a subspace, you must check that:
  • The zero vector is within the set, helping maintain the identity aspect of addition.
  • It is closed under vector addition—meaning any two vectors added together still reside in the subspace.
  • It is closed under scalar multiplication—ensuring that scaling a vector by any scalar keeps it within the subspace.
By evaluating if these conditions are met, one can determine if a given set, like the set of points inside and on a unit circle, can be deemed a subspace of a larger vector space like \( \mathbb{R}^2 \). However, as shown in our original exercise, if even one condition is not satisfied, such as closure under scalar multiplication, the set cannot be a subspace.
Scalar Multiplication
Scalar multiplication is an operation that involves multiplying a vector by a scalar, resulting in a new vector. In the context of vector spaces, this operation must keep the resultant vector within the same space for closure. Consider a vector \( \mathbf{v} \) in a vector space \( V \) and a scalar \( c \). Scalar multiplication demands that \( c \mathbf{v} \) is also within \( V \). This property is crucial in preserving the structure of a vector space or its subspaces. In the specific example of the unit circle, while the vector \( \begin{bmatrix} 1 \ 0 \end{bmatrix} \) is within the set \( H \), multiplying it by a scalar such as 2 results in \( \begin{bmatrix} 2 \ 0 \end{bmatrix} \), thus shifting it outside the bounds defined by \( H \). This failure to remain within the unit circle illustrates that \( H \) is not closed under scalar multiplication, disqualifying it as a subspace. Such an example readily shows the importance of this condition in defining subspace behavior within vector spaces.

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

In Exercises 29 and \(30, V\) is a nonzero finite-dimensional vector space, and the vectors listed belong to \(V\) . Mark each statement True or False. Justify each answer. (These questions are more difficult than those in Exercises 19 and \(20 .\) ) a. If there exists a set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) that spans \(V,\) then \(\operatorname{dim} V \leq p .\) b. If there exists a linearly independent set \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) in \(V,\) then \(\operatorname{dim} V \geq p\) c. If \(\operatorname{dim} V=p,\) then there exists a spanning set of \(p+1\) vectors in \(V .\)

In \(\mathbb{P}_{2},\) find the change-of-coordinates matrix from the basis \(\mathcal{B}=\left\\{1-3 t^{2}, 2+t-5 t^{2}, 1+2 t\right\\}\) to the standard basis. Then write \(t^{2}\) as a linear combination of the polynomials in \(\mathcal{B}\) .

Exercises 31 and 32 reveal an important connection between linear independence and linear transformations and provide practice using the definition of linear dependence. Let \(V\) and \(W\) be vector spaces, let \(T : V \rightarrow W\) be a linear transformation, and let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) be a subset of \(V .\) Suppose that \(T\) is a one-to-one transformation, so that an equation \(T(\mathbf{u})=T(\mathbf{v})\) always implies \(\mathbf{u}=\mathbf{v} .\) Show that if the set of images \(\left\\{T\left(\mathbf{v}_{1}\right), \ldots, T\left(\mathbf{v}_{p}\right)\right\\}\) is linearly dependent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set (because in this case the set of images cannot be linearly dependent).

In Exercises 17 and 18 , \(A\) is an \(m \times n\) matrix. Mark each statement True or False. Justify each answer. a. If \(B\) is any echelon form of \(A\) , then the pivot columns of \(B\) form a basis for the column space of \(A\) . b. Row operations preserve the linear dependence relations among the rows of \(A\) . c. The dimension of the null space of \(A\) is the number of columns of \(A\) that are not pivot columns. d. The row space of \(A^{T}\) is the same as the column space of \(A .\) e. If \(A\) and \(B\) are row equivalent, then their row spaces are the same.

In Exercises \(7-12\) , assume the signals listed are solutions of the given difference equation. Determine if the signals form a basis for the solution space of the equation. Justify your answers using appropriate theorems. $$ 2^{k}, 4^{k},(-5)^{k} ; y_{k+3}-y_{k+2}-22 y_{k+1}+40 y_{k}=0 $$

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.