/*! 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 62 Show that if \(A\) is an \(m \ti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 62

Show that if \(A\) is an \(m \times n\) matrix, then \(\operatorname{ker}(A)\) is a subspace of \(\mathbb{R}^{n} .\)

Short Answer

Expert verified
The kernel of matrix A is a subspace of \( \textbf{R}^{n} \) because it contains the zero vector and is closed under addition and scalar multiplication.

Step by step solution

01

- Define the Kernel of Matrix

The kernel of a matrix A, denoted as \(\text{ker}(A)\), is the set of all vectors \(\textbf{v} \) in \(\textbf{R}^{n} \) such that \(\text{A}\textbf{v} = \textbf{0}\).
02

- Check Zero Vector

A subspace must contain the zero vector. Check that the zero vector \( \textbf{0} \) belongs to \( \text{ker}(A) \): \(\text{A}\textbf{0} = \textbf{0} \). Thus, \( \textbf{0} \) is in the kernel.
03

- Check Closure Under Addition

To verify that \( \text{ker}(A) \) is closed under addition, take any two vectors \( \textbf{v}_1 \) and \( \textbf{v}_2 \) in \( \text{ker}(A) \). Show \( \text{A}(\textbf{v}_1 + \textbf{v}_2) = \text{A}\textbf{v}_1 + \text{A}\textbf{v}_2 = \textbf{0} + \textbf{0} = \textbf{0} \). Therefore, the sum \( \textbf{v}_1 + \textbf{v}_2 \) is in the kernel.
04

- Check Closure Under Scalar Multiplication

Check if \( \text{ker}(A) \) is closed under scalar multiplication. Take any vector \( \textbf{v} \) in \( \text{ker}(A) \) and any scalar \( c \). Show \( \text{A}(c\textbf{v}) = c\text{A}\textbf{v} = c\textbf{0} = \textbf{0} \). Hence, \( c\textbf{v} \) is in the kernel.
05

- Conclude

Since \( \text{ker}(A) \) contains the zero vector, is closed under addition, and is closed under scalar multiplication, it satisfies the conditions to be a subspace of \( \textbf{R}^{n} \).

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.

subspace
A subspace is a set of vectors that follows certain rules within a larger vector space. Specifically, it needs to satisfy three main conditions: it must include the zero vector, be closed under addition, and be closed under scalar multiplication.

When we check if a set is a subspace, we are ensuring that it behaves nicely under these basic operations. This ensures any operation within this set does not lead us outside of it.
linear transformation
A linear transformation is a function between vector spaces that preserves the operations of vector addition and scalar multiplication. This means if you apply the function to a combination of vectors, you can instead combine the results of applying the function to each vector individually and get the same output.
In mathematical terms, for a function T, this means:
  • T(a + b) = T(a) + T(b) for vectors a and b

  • T(ca) = cT(a) for any scalar c and vector a
Linear transformations are foundational in understanding the behavior of systems of linear equations and matrices.
zero vector
The zero vector is the unique vector in any vector space that has all components equal to zero. For an n-dimensional vector space, the zero vector is written as \( \textbf{0} = [0, 0, ..., 0] \).

The zero vector is crucial in defining a subspace because it needs to be part of any subspace by definition. Without the zero vector, the set cannot be considered a subspace.
closure under addition
Closure under addition is a property that says if you take any two vectors in a set, their sum must also be in that same set. This is key to ensuring the set of vectors acts like a subspace.
For instance, if vectors \textbf{u}\ and \textbf{v}\ are in the kernel of a matrix A, the sum \textbf{u} + \textbf{v}\ must also be in the kernel. So, \(A(\textbf{u} + \textbf{v}) = \textbf{0}\), meaning that the kernel is closed under addition.
closure under scalar multiplication
Closure under scalar multiplication ensures that if you multiply a vector in a set by any scalar, the resulting vector will still be in that set. This is essential for confirming something is a subspace.

For example, if \textbf{v}\ is in the kernel of matrix A, and c is any scalar, then c \textbf{v}\ must also be in the kernel. In mathematical terms, \(A(c\textbf{v}) = cA(\textbf{v}) = \textbf{0}\). This demonstrates that the kernel remains closed under scalar multiplication, fulfilling another criterion for being a subspace.

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

Here are some matrices. Label according to whether they are symmetric, skew symmetric, or orthogonal. $$ \begin{array}{l} \text { (a) }\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right] \\ \text { (b) }\left[\begin{array}{rrr} 1 & 2 & -3 \\ 2 & 1 & 4 \\ -3 & 4 & 7 \end{array}\right] \\ \text { (c) }\left[\begin{array}{rrr} 0 & -2 & -3 \\ 2 & 0 & -4 \\ 3 & 4 & 0 \end{array}\right] \end{array} $$

The wind blows from West to East at a speed of 50 miles per hour and an airplane which travels at 400 miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?

Suppose A is an \(m \times n\) matrix and \(B\) is an \(n \times p\) matrix. Show that $$ \operatorname{dim}(\operatorname{ker}(A B)) \leq \operatorname{dim}(\operatorname{ker}(A))+\operatorname{dim}(\operatorname{ker}(B)) $$ Hint: Consider the subspace, \(B\left(\mathbb{R}^{p}\right) \cap \operatorname{ker}(A)\) and suppose a basis for this subspace is \(\left\\{\vec{w}_{1}, \cdots, \vec{w}_{k}\right\\} .\) Now suppose \(\left\\{\vec{u}_{1}, \cdots, \vec{u}_{r}\right\\}\) is a basis for \(\operatorname{ker}(B) .\) Let \(\left\\{\vec{z}_{1}, \cdots, \vec{z}_{k}\right\\}\) be such that \(B \vec{z}_{i}=\vec{w}_{i}\) and argue that $$ \operatorname{ker}(A B) \subseteq \operatorname{span}\left\\{\vec{u}_{1}, \cdots, \vec{u}_{r}, \vec{z}_{1}, \cdots, \vec{z}_{k}\right\\} $$

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $$ \left[\begin{array}{r} 2 \\ 3 \\ 1 \\ -3 \end{array}\right],\left[\begin{array}{r} -5 \\ -6 \\ 0 \\ 3 \end{array}\right],\left[\begin{array}{r} -1 \\ -2 \\ 1 \\ 3 \end{array}\right],\left[\begin{array}{r} -1 \\ -2 \\ 0 \\ 4 \end{array}\right] $$

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $$ \left[\begin{array}{r} 1 \\ 4 \\ -2 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 5 \\ -3 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 7 \\ -5 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 5 \\ -2 \\ 1 \end{array}\right] $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

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.