/*! 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 12 We often write vectors in \(\mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 12

We often write vectors in \(\mathbb{R}^{n}\) as rows. Suppose that \(U=\operatorname{span}\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \ldots, \mathbf{x}_{k}\right\\}\) where each \(\mathbf{x}_{i}\) is in \(\mathbb{R}^{n}\). If \(A\) is an \(m \times n\) matrix and \(A \mathbf{x}_{i}=\mathbf{0}\) for each \(i\), show that \(A \mathbf{y}=\mathbf{0}\) for every vector \(\mathbf{y}\) in \(U\)

Short Answer

Expert verified
Vectors in the span of \(\{\mathbf{x}_i\}\) map to \(\mathbf{0}\) by \(A\).

Step by step solution

01

Understanding the components

We are given that vectors \(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\) are in \(\mathbb{R}^n\) and they span a subspace \(U\). Also, \(A\) is an \(m \times n\) matrix such that the product of \(A\) with any of these vectors results in the zero vector \(\mathbf{0}\) in \(\mathbb{R}^m\). We need to show this property holds for any vector \(\mathbf{y}\) in the span of these vectors as well.
02

Defining Vector \( \mathbf{y} \) in \( U \)

Since \( U \) is the span of \(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k\), any vector \(\mathbf{y} \in U\) can be expressed as a linear combination: \(\mathbf{y} = c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k\), where \( c_1, c_2, \ldots, c_k \) are scalars.
03

Multiply by matrix \( A \)

Compute the product \( A\mathbf{y} \). Substituting the expression for \(\mathbf{y}\), we have: \[ A \mathbf{y} = A(c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k) \].
04

Using Matrix Linearity

Apply the distributive property and linearity of matrix multiplication: \[ A \mathbf{y} = c_1 A \mathbf{x}_1 + c_2 A \mathbf{x}_2 + \cdots + c_k A \mathbf{x}_k \].
05

Substituting \(A \mathbf{x}_i = \mathbf{0} \)

Given that \(A \mathbf{x}_i = \mathbf{0} \) for each \(i\), substitute \(\mathbf{0}\) into the expression: \[ A \mathbf{y} = c_1 \mathbf{0} + c_2 \mathbf{0} + \cdots + c_k \mathbf{0} \].
06

Final simplification

Since all terms are scalar multiples of the zero vector, they simplify to the zero vector: \[ A \mathbf{y} = \mathbf{0} + \mathbf{0} + \cdots + \mathbf{0} = \mathbf{0} \].
07

Conclusion

Thus, for any vector \( \mathbf{y} \) in \( U \), it holds that \( A\mathbf{y} = \mathbf{0} \). The property \(A\mathbf{x}_i = \mathbf{0} \) extends to any vector in the span \( U \).

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 Spaces
In linear algebra, the concept of vector spaces is fundamental. A vector space is a collection of vectors that can be added together and multiplied by scalars to produce another vector within the same space. This is crucial because it provides a framework for many operations that we perform with vectors. In our given problem, the span of vectors \( \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k \) forms a subspace \( U \) within the vector space \( \mathbb{R}^n \).
  • **Span:** The span of a set of vectors is all possible linear combinations of those vectors. For instance, any vector \( \mathbf{y} \) in \( U \) can be expressed using the vectors \( \mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_k \).
  • **Subspace:** A subspace is a subset of a vector space that is itself a vector space under the operations of vector addition and scalar multiplication. In this case, \( U \) is a subspace of \( \mathbb{R}^n \).
The operations allowed in a vector space, such as addition and scalar multiplication, are critical in understanding how vectors interact with matrices, leading into our discussion on matrix multiplication.
Matrix Multiplication
Matrix multiplication is a procedure where two matrices are multiplied to produce another matrix. In our problem, we're particularly interested in multiplying the matrix \( A \) by vectors from our space \( U \). Understanding how this operation works is key to solving the exercise.
  • **Linearity:** One essential property of matrix multiplication is its linearity. This means that \( A(c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k) \) can be expanded to \( c_1A\mathbf{x}_1 + c_2A\mathbf{x}_2 + \ldots + c_kA\mathbf{x}_k \).
  • **Zero Product:** Given the problem, we are told that \( A\mathbf{x}_i = \mathbf{0} \) for each vector \( \mathbf{x}_i \) in the subspace. This simplifies each term of our expanded expression to zero, demonstrating that \( A\mathbf{y} = \mathbf{0} \) for any vector \( \mathbf{y} \) in \( U \).
This demonstrates how efficiently operations on matrices facilitate manipulations and proofs in linear algebra, particularly when validating results across entire subspaces.
Linear Combinations
Linear combinations are expressions that involve adding together multiple vectors, each scaled by a certain factor. This concept is crucial to our understanding of vector spaces and the exercise at hand. When working in a vector space, we are often interested in how vectors can be combined using scalars. This allows us to construct new vectors, like the vector \( \mathbf{y} \) in the span \( U \) from the problem.
  • **Building New Vectors:** In vector spaces, any vector can frequently be represented as a linear combination. For the subspace \( U \), any \( \mathbf{y} \) is \( c_1\mathbf{x}_1 + c_2\mathbf{x}_2 + \cdots + c_k\mathbf{x}_k \).
  • **Role in Proofs:** In our problem, this representation of vectors as linear combinations allows us to break down the multiplication \( A\mathbf{y} \) step by step, using properties like linearity to conclude the result \( \mathbf{0} \).
Understanding linear combinations allows us to see how vectors interrelate within their space, providing a deeper insight into the structures and results of linear transformations in algebra.

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

We often write vectors in \(\mathbb{R}^{n}\) as rows. Let \(P\) denote an invertible \(n \times n\) matrix. If \(\lambda\) is a number, show that $$ E_{\lambda}\left(P A P^{-1}\right)=\left\\{P \mathbf{x} \mid \mathbf{x} \text { is in } E_{\lambda}(A)\right\\} $$ for each \(n \times n\) matrix \(A\).

Let \(\mathbf{v}\) denote a nonzero vector in \(\mathbb{R}^{n}\). a. Show that \(P=\left\\{\mathbf{x}\right.\) in \(\left.\mathbb{R}^{n} \mid \mathbf{x} \cdot \mathbf{v}=0\right\\}\) is a subspace of \(\mathbb{R}^{n}\) b. Show that \(\mathbb{R} \mathbf{v}=\\{t \mathbf{v} \mid t\) in \(\mathbb{R}\\}\) is a subspace of \(\mathbb{R}^{n}\) c. Describe \(P\) and \(\mathbb{R} \mathbf{v}\) geometrically when \(n=3\).

Find the least squares approximating function of the form \(r_{0}+r_{1} x^{2}+r_{2} \sin \frac{\pi x}{2}\) for each of the following sets of data pairs. a. (0,3),(1,0),(1,-1),(-1,2) b. \(\left(-1, \frac{1}{2}\right),(0,1),(2,5),(3,9)\)

}\( satisfying the following four conditions: \)A A^{\\#} A=A ; A^{\鈥� # If \(A\) is an \(m \times n\) matrix, it can be proved that there exists a unique \(n \times m\) matrix \(A^{\\#}\) satisfying the following four conditions: \(A A^{\\#} A=A ; A^{\\#} A A^{\\#}=A^{\\#} ; A A^{\\#}\) and \(A^{\\#} A\) are symmetric. The matrix \(A^{\\#}\) is called the generalized inverse of \(A\), or the Moore-Penrose inverse. a. If \(A\) is square and invertible, show that \(A^{\\#}=A^{-1}\). b. If \(\operatorname{rank} A=m,\) show that \(A^{\\#}=A^{T}\left(A A^{T}\right)^{-1}\). c. If \(\operatorname{rank} A=n,\) show that \(A^{\\#}=\left(A^{T} A\right)^{-1} A^{T}\).

We often write vectors in \(\mathbb{R}^{n}\) as rows. Let \(U\) be a subspace of \(\mathbb{R}^{n},\) and let \(\mathbf{x}\) be a vector in \(\mathbb{R}^{n}\). a. If \(a \mathbf{x}\) is in \(U\) where \(a \neq 0\) is a number, show that \(\mathbf{x}\) is in \(U\) b. If \(\mathbf{y}\) and \(\mathbf{x}+\mathbf{y}\) are in \(U\) where \(\mathbf{y}\) is a vector in \(\mathbb{R}^{n}\), show that \(\mathbf{x}\) is in \(U\)
See all solutions

Recommended explanations on Economics Textbooks

Taxation Economics

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.