/*! 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 35 Let \(V\) and \(W\) be vector sp... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(V\) and \(W\) be vector spaces, and let \(T : V \rightarrow W\) be a linear transformation. Given a subspace \(U\) of \(V,\) let \(T(U)\) denote the set of all images of the form \(T(\mathbf{x}),\) where \(\mathbf{x}\) is in \(U .\) Show that \(T(U)\) is a subspace of \(W .\)

Short Answer

Expert verified
\(T(U)\) is a subspace of \(W\) because it is non-empty and closed under addition and scalar multiplication.

Step by step solution

01

Understanding Subspaces

To show that a set is a subspace, we need to prove it has three properties: it is non-empty, it is closed under addition, and it is closed under scalar multiplication.
02

Non-emptiness of \(T(U)\)

Since \(U\) is a subspace of \(V\), it includes the zero vector \(\mathbf{0}_V\). Since \(T\) is a linear transformation, \(T(\mathbf{0}_V) = \mathbf{0}_W\). Therefore, \(\mathbf{0}_W\) is in \(T(U)\), proving it is non-empty.
03

Closure Under Addition

Consider any two vectors \(\mathbf{a}, \mathbf{b} \in U\). Then \(T(\mathbf{a})\) and \(T(\mathbf{b})\) are in \(T(U)\). Since \(T\) is a linear transformation, \(T(\mathbf{a} + \mathbf{b}) = T(\mathbf{a}) + T(\mathbf{b})\). Hence, \(T(\mathbf{a}) + T(\mathbf{b})\) is in \(T(U)\), proving closure under addition.
04

Closure Under Scalar Multiplication

Consider any vector \(\mathbf{x} \in U\) and any scalar \(c \in \mathbb{R}\). \(T(\mathbf{x})\) is in \(T(U)\), and since \(T\) is linear, \(T(c\mathbf{x}) = cT(\mathbf{x})\). Therefore, \(cT(\mathbf{x})\) is in \(T(U)\), proving closure under scalar multiplication.

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 fundamental concept in linear algebra. It is a collection of objects called vectors, which can be added together and multiplied by scalars (numbers), and still remain within the collection. In a more formal sense, a vector space \( V \) over a field \( F \) (such as the real numbers \( \mathbb{R} \)) has the following properties:

  • Closure under addition: The sum of any two vectors in \( V \) is also in \( V \).
  • Closure under scalar multiplication: The product of any vector in \( V \) and a scalar from \( F \) is also in \( V \).
  • Identity element of addition: There exists a zero vector \( \mathbf{0} \) in \( V \) such that for any vector \( \mathbf{v} \) in \( V \), \( \mathbf{v} + \mathbf{0} = \mathbf{v} \).
  • Inverses for addition: For every vector \( \mathbf{v} \) in \( V \), there is a vector \( -\mathbf{v} \) such that \( \mathbf{v} + (-\mathbf{v}) = \mathbf{0} \).
A vector space can host many different mathematical phenomena, making it a powerful tool in understanding the nature of linear transformations.
Subspace
A subspace is like a mini vector space contained within a larger vector space. It must satisfy all the properties of a vector space within its larger encompassing space. For a set \( U \) to be a subspace of a vector space \( V \), it needs to meet these criteria:

  • Non-empty: It must include the zero vector from \( V \).
  • Closed under addition: If \( \mathbf{u} \) and \( \mathbf{v} \) are in \( U \), then \( \mathbf{u} + \mathbf{v} \) is also in \( U \).
  • Closed under scalar multiplication: If \( \mathbf{u} \) is in \( U \), then \( c\mathbf{u} \) is also in \( U \) for any scalar \( c \).
Subspaces are important because they allow us to handle smaller, more manageable pieces that share the vector space's overall properties, and this is especially useful when working with linear transformations.
Closure Properties
Closure properties are an essential aspect of understanding vector spaces and subspaces. They ensure that the operations we perform remain within the set we are examining. There are two main closure properties to consider:

Closure Under Addition: When two vectors from a subspace (or vector space) are added together, their sum must also be in that subspace. For example, if \( \mathbf{a} \) and \( \mathbf{b} \) are in subspace \( U \), then \( \mathbf{a} + \mathbf{b} \) must also be in \( U \). Linear transformations also respect this property, which is why \( T(\mathbf{a}) + T(\mathbf{b}) \) remains in \( T(U) \).

Closure Under Scalar Multiplication: When a vector is multiplied by a scalar, the resulting vector must also lie within the subspace. This means if \( c \) is any scalar and \( \mathbf{x} \) is a vector in subspace \( U \), then \( c \mathbf{x} \) must also be in \( U \). For linear transformations, this property is preserved as \( T(c\mathbf{x}) = cT(\mathbf{x}) \).

These closures are critical for the stability and consistency of vector operations and ensure that the subspace structure remains intact.
Scalar Multiplication
Scalar multiplication in vector spaces is an operation that involves multiplying a vector by a scalar (a single number, often real or complex). This operation should not alter the vector's membership in the space. Here’s how it works:

When you take a scalar \( c \) and a vector \( \mathbf{x} \), scalar multiplication results in another vector \( c\mathbf{x} \) within the same space. If \( \mathbf{x} \) belongs to a subspace \( U \) of a vector space \( V \), then \( c\mathbf{x} \) also remains in \( U \).

  • The direction of the vector remains the same if the scalar is positive, but its magnitude changes according to the absolute value of the scalar.
  • If the scalar is negative, the vector not only changes magnitude but also direction.
  • Scalar multiplication with zero results in the zero vector, effectively "erasing" any directional information.
This concept is crucial in understanding how linear transformations distribute across scalar multiplication, maintaining the vector space's integrity by ensuring outcomes remain within the confines of their respective subspaces.

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

Rank 1 matrices are important in some computer algorithms and several theoretical contexts, including the singular value decomposition in Chapter \(7 .\) It can be shown that an \(m \times n\) matrix \(A\) has rank 1 if and only if it is an outer product; that is, \(A=\mathbf{u v}^{T}\) for some \(\mathbf{u}\) in \(\mathbb{R}^{m}\) and \(\mathbf{v}\) in \(\mathbb{R}^{n} .\) Exercises \(31-33\) suggest why this property is true. Let \(\mathbf{u}=\left[\begin{array}{l}{1} \\ {2}\end{array}\right] .\) Find \(\mathbf{v}\) in \(\mathbb{R}^{3}\) such that \(\left[\begin{array}{ccc}{1} & {-3} & {4} \\ {2} & {-6} & {8}\end{array}\right]=\mathbf{u v}^{T}\)

If \(A\) is a \(6 \times 4\) matrix, what is the smallest possible dimension of Nul \(A ?\)

In Exercises 13 and 14 , assume that \(A\) is row equivalent to \(B .\) Find bases for Nul \(A\) and \(\operatorname{Col} A .\) $$ A=\left[\begin{array}{rrrr}{-2} & {4} & {-2} & {-4} \\ {2} & {-6} & {-3} & {1} \\ {-3} & {8} & {2} & {-3}\end{array}\right], B=\left[\begin{array}{llll}{1} & {0} & {6} & {5} \\ {0} & {2} & {5} & {3} \\\ {0} & {0} & {0} & {0}\end{array}\right] $$

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

A small remote village receives radio broadcasts from two radio stations, a news station and a music station. Of the listeners who are tuned to the news station, 70\(\%\) will remain listening to the news after the station break that occurs each half hour, while 30\(\%\) will switch to the music station at the station break. Of the listeners who are tuned to the music station, 60\(\%\) will switch to the news station at the station break, while 40\(\%\) will remain listening to the music. Suppose everyone is listening to the news at \(8 : 15\) A.M. a. Give the stochastic matrix that describes how the radio listeners tend to change stations at each station break. Label the rows and columns. b. Give the initial state vector. c. What percentage of the listeners will be listening to the music station at \(9 : 25\) A.M. (after the station breaks at \(8 : 30\) and \(9 : 00\) A.M. \() ?\)

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.