/*! 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 24 Let \(T: V \rightarrow V\) be a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 24

Let \(T: V \rightarrow V\) be a linear transformation of a vector space \(V\) into itself, and let \(\lambda\) be a scalar. Prove that \(\\{\mathbf{v} \in V \mid T(\mathbf{v})=\lambda \mathbf{v}\\}\) is a subspace of \(V\).

Short Answer

Expert verified
The set \( \{\mathbf{v} \in V \mid T(\mathbf{v}) = \lambda \mathbf{v} \} \) is a subspace of \( V \).

Step by step solution

01

Define the Set

Let \( W = \{\mathbf{v} \in V \mid T(\mathbf{v}) = \lambda \mathbf{v} \} \). We need to prove that \( W \) is a subspace of \( V \). A subset of a vector space is a subspace if it is closed under addition and scalar multiplication, and contains the zero vector.
02

Verify the Zero Vector in W

To check if the zero vector \( \mathbf{0} \) is in \( W \), apply the transformation: \( T(\mathbf{0}) = 0 = \lambda \cdot \mathbf{0} \). Since \( T(\mathbf{0}) = \lambda \cdot \mathbf{0} \), the zero vector is indeed in \( W \).
03

Check Closure Under Addition

Take any \( \mathbf{u}, \mathbf{v} \in W \). By definition, \( T(\mathbf{u}) = \lambda \mathbf{u} \) and \( T(\mathbf{v}) = \lambda \mathbf{v} \). Consider \( \mathbf{u} + \mathbf{v} \). Then:\[ T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) = \lambda \mathbf{u} + \lambda \mathbf{v} = \lambda (\mathbf{u} + \mathbf{v}) \]Therefore, \( \mathbf{u} + \mathbf{v} \in W \).
04

Check Closure Under Scalar Multiplication

Take any \( \mathbf{v} \in W \) and any scalar \( c \). Since \( T(\mathbf{v}) = \lambda \mathbf{v} \), consider \( c\mathbf{v} \). Then:\[ T(c\mathbf{v}) = cT(\mathbf{v}) = c(\lambda \mathbf{v}) = \lambda (c\mathbf{v}) \]Thus, \( c\mathbf{v} \in W \).
05

Conclusion

Since \( W \) contains the zero vector, is closed under addition, and closed under scalar multiplication, \( W \) is a subspace of \( V \).

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 91影视!

Key Concepts

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

Vector Space
Imagine a vector space as a playground for vectors. It is a collection of objects called vectors, along with two operations: vector addition and scalar multiplication.
  • Vector addition allows you to combine two vectors, producing another vector in the same space.
  • Scalar multiplication lets you stretch or shrink a vector using a number (a scalar).
Vector spaces are everywhere in mathematics and have specific properties that make them very structured:
  • There is always a zero vector, which acts as an additive identity.
  • The space is closed under addition and scalar multiplication, meaning that combining vectors or scaling them stays inside the space.
This concept of closure ensures that every operation inside the vector space retains its vector nature. Understanding the basics of vector spaces helps to comprehend more complex ideas like linear transformations and subspace.
Subspace
A subspace is like a smaller playground within a bigger vector space. It itself forms a vector space and obeys the same rules.
A subspace of a vector space is a set of vectors that satisfies the following:
  • Contains the zero vector.
  • Closed under addition: If you take any two vectors in this set, their sum is also in the set.
  • Closed under scalar multiplication: If you multiply any vector in the subspace by any scalar, the result still lies in the subspace.
For the exercise given, we proved that the specific set, where each vector satisfies the condition given by the linear transformation, is a subspace. As long as it meets these three criteria, we can be assured that the set maintains all the necessary properties of being a subspace.
Scalar Multiplication
Scalar multiplication is like changing the size of a vector without altering its direction.
If you imagine a vector as an arrow, scalar multiplication stretches or compresses the arrow. Consider multiplying a vector by a positive scalar 鈥� it increases or decreases its length while maintaining its original direction.
  • If the scalar is negative, the vector also flips direction but preserves its magnitude change.
  • If the scalar is zero, the vector "disappears" into the zero vector.
This operation is vital in checking if a set of vectors forms a subspace, as multiplying any vector in a prospective subspace by a scalar should also result in a vector still within that subspace. In the solution to our exercise, scalar multiplication supports determining if the set is a subspace. It's a key property that ensures the integrity of the vector space's structure when vectors are scaled.
Closure Properties
Closure properties are like the safety rules on the playground, ensuring you stay within the limits.
For a set of vectors to be considered a vector space or subspace, it must be closed under two main operations: addition and scalar multiplication.
  • Closed under addition means that adding any two vectors in the set results in another vector that still belongs to the set.
  • Closed under scalar multiplication means any vector in the set, when multiplied by a scalar, remains in the set.
These properties are essential in our exercise because showing closure under addition and scalar multiplication proves that the set {\(\mathbf{v}\)} is indeed a subspace.When these closure conditions are satisfied, you can confidently use vectors in any operations knowing they stay within the defined space, maintaining the structure and rules of the vector space. This ensures consistency and reliability in mathematical computations.

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

Prove that similar square matrices have the same eigenvalues with the same algebraic multiplicities.

Find the eigenvalues \(\lambda_{1}\) and the Eorresponding eigenvectors \(\mathrm{v}_{i}\), of the given matrix Li and also find an invertible matrix \(C\) and a fiagonal matrix \(D\) such that \(D=C^{-1} A C\). $$ A=\left[\begin{array}{rrr} -2 & 0 & -1 \\ 0 & 2 & 0 \\ 3 & 0 & 2 \end{array}\right] $$

For a square matrix \(A\), the \(M A T L A B\) command Eg(A) produces the eigenvalues (both real and complex) of \(A\). The command \([\mathrm{V}, \mathrm{D}]=\operatorname{eig}(\mathrm{A})\) produces a matrix \(V\) whose column vectors are (perhaps complex) eigenvectors of \(A\) and a diagonal matrix \(D\) whose entry \(d_{i i}\) is the eigenvalue for the eigenvector in the ith column of V. In Exercises 48-51, use either MATLAB or the in routine MATCOMP in LINTEK to find the real eigenvalues and corresponding eigenvectors of the given matrix. $$ \left[\begin{array}{rrrr} 0 & 0 & -2 & 2 \\ -3 & 4 & -3 & 3 \\ -4 & 0 & 2 & 4 \\ -2 & 0 & 2 & 4 \end{array}\right] $$

Determine whether the given matrix is diagonalizable. $$ \left[\begin{array}{rrrr} 3 & 2 & 5 & 1 \\ 2 & 0 & 2 & 6 \\ 5 & 2 & 7 & -1 \\ 1 & 6 & -1 & 3 \end{array}\right] $$

Mark each of the following True or False. a. Every square matrix has real eigenvalues. b. Every \(n \times n\) matrix has \(n\) distinct (possibly complex) eigenvalues. c. Every \(n \times n\) matrix has \(n\) not necessarily distinct and possibly complex eigenvalues. d. There can be only one eigenvalue associated with an eigenvector of a linear transformation. e. There can be only one eigenvector associated with an eigenvalue of a linear transformation. f. If \(\mathbf{v}\) is an eigenvector of a matrix \(A\), then \(\mathbf{v}\) is an eigenvector of \(A+c I\) for all scalars \(c\). g. If \(\lambda\) is an eigenvalue of a matrix \(A\), then \(\lambda\) is an eigenvalue of \(A+c I\) for all scalars \(c\). h. If \(\mathbf{v}\) is an eigenvector of an invertible matrix \(A\), then \(c v\) is an eigenvector of \(A^{-1}\) for all nonzero scalars \(c\). i. Every vector in a vector space \(V\) is an eigenvector of the identity transformation of \(V\) into \(V\). j. Every nonzero vector in a vector space \(V\) is an eigenvector of the identity transformation of \(V\) into \(V\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation
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.