/*! 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 28 Exercises \(27-29\) concern an \... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 28

Exercises \(27-29\) concern an \(m \times n\) matrix \(A\) and what are often called the fundamental subspaces determined by \(A .\) Justify the following equalities: a. \(\operatorname{dim} \mathrm{Row} A+\operatorname{dim} \mathrm{Nul} A=n\) Number of columns of \(A\) b. \(\operatorname{dim} \operatorname{Col} A+\operatorname{dim} \mathrm{Nul} A^{T}=m\) Number of rows of \(A\)

Short Answer

Expert verified
The equalities follow from the rank-nullity theorem.

Step by step solution

01

Understand the Problem

We need to demonstrate two mathematical equalities related to the dimensions of fundamental subspaces of a matrix \(A\). The fundamental subspaces include the row space, column space, null space, and null space of the transpose.
02

Define the Terms

The row space of a matrix \(A\), denoted \(\text{Row } A\), is the span of the row vectors of \(A\). The null space, \(\text{Nul } A\), consists of all solutions to the equation \(A\mathbf{x} = 0\). The column space, \(\text{Col } A\), is the span of the columns of \(A\). The null space of the transpose, \(\text{Nul } A^T\), consists of all solutions to \(A^T\mathbf{y} = 0\).
03

Apply the Rank-Nullity Theorem to Row Space and Null Space

For part (a), we use the rank-nullity theorem which states that \( \text{dim Row } A + \text{dim Nul } A = n \), where \(n\) is the number of columns of \(A\). The rank of \(A\), which is \(\text{dim Row } A\), plus the nullity of \(A\), which is \(\text{dim Nul } A\), should equal the total number of columns.
04

Apply the Rank-Nullity Theorem to Column Space and Null Space of Transpose

For part (b), again apply the rank-nullity theorem but to the transpose of \(A\). This states that \( \text{dim Col } A + \text{dim Nul } A^T = m \), where \(m\) is the number of rows of \(A\). The rank of the transpose, which is \(\text{dim Col } A\), plus the nullity of the transpose, \(\text{dim Nul } A^T\), equals to the number of rows \(m\).
05

Conclusion of the Equalities

By applying the rank-nullity theorem, we have shown that the dimensions of the row space and null space together equal the number of columns, and the dimensions of the column space and the null space of the transpose equal the number of rows.

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Key Concepts

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

Row Space
The row space of a matrix \(A\), also known as the "row rank," is the set of all possible linear combinations of its row vectors. This space is crucial because it tells us about the linear relations between the rows of \(A\). If you're imagining a two-dimensional plane, the row space is like a flat surface hovering somewhere in the multi-dimensional space.
  • If all rows were linear combinations of each other, the dimension of the row space would be 1.
  • If the rows span a plane, then the dimension would be 2.
  • This dimension is also known as the "rank" of the matrix and represents the maximum number of linearly independent row vectors in \(A\).
Understanding the row space helps us to check for row dependencies, and it is crucial in matrix operations such as solving linear equations.
Null Space
The null space of a matrix \(A\), denoted as \(\text{Nul } A\), is the set of all vectors \(\mathbf{x}\) such that \(A\mathbf{x} = 0\). In simple terms, it represents all solutions to the homogeneous equation formed by \(A\).
  • One can think of the null space as capturing the 'hidden' directions in space where multiplying by \(A\) results in a zero vector.
  • The dimension of the null space is called the 'nullity'.
  • If the nullity is zero, it implies that there are no such hidden directions and that \(A\) is invertible (for a square matrix).
Understanding the null space is essential when you want to understand the number of solutions a system of equations can have. It reveals how \(A\) "crunches" vectors to zero and is a pivotal concept in linear transformations.
Rank-Nullity Theorem
The Rank-Nullity Theorem is a fundamental principle in linear algebra that connects all these subspaces of a matrix into one simple equation.
  • For any matrix \(A\) of size \(m \times n\), the theorem states: \( \text{dim Row } A + \text{dim Nul } A = n \).
  • This means the rank (dimension of row space) plus the nullity (dimension of the null space) always adds up to the number of columns \(n\).
  • For the transpose of \(A\), it relates the column space and the null space of the transpose: \( \text{dim Col } A + \text{dim Nul } A^T = m \), the total number of rows.
This wonderful theorem gives us insight into the structure of a matrix and how these subspaces interact. It's essentially a "balance equation" for the dimensions, providing a check for consistency in solving systems of equations.
Column Space
The column space of a matrix \(A\), sometimes simply called the "image," is the set of all vectors that can be expressed as linear combinations of the column vectors of \(A\). Each column can be considered as a point extending in space from the origin.
  • The dimension of the column space, termed the rank of \(A\), measures the number of linearly independent columns.
  • If the rank equals the number of rows, the column vectors span the whole space; this is a special case for square matrices where the matrix is considered to be "full rank" and thus invertible.
  • The column space is crucial for solving linear systems because it tells us about the potential outputs or results that can be achieved via the transformation described by \(A\).
Visualizing the column space gives insight into how the transformation affects the entire vector space, forming an essential component of matrix transformations.

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Most popular questions from this chapter

Determine if \(P=\left[\begin{array}{cc}{1} & {.2} \\ {0} & {.8}\end{array}\right]\) is a regular stochastic matrix.

Let \(\mathcal{B}=\left\\{\mathbf{b}_{1}, \mathbf{b}_{2}\right\\}\) and \(\mathcal{C}=\left\\{\mathbf{c}_{1}, \mathbf{c}_{2}\right\\}\) be bases for \(\mathbb{R}^{2} .\) In each exercise, find the change-of-coordinates matrix from \(\mathcal{B}\) to \(\mathcal{C}\) and the change-of-coordinates matrix from \(\mathcal{C}\) to \(\mathcal{B} .\) \(\mathbf{b}_{1}=\left[\begin{array}{r}{-1} \\ {8}\end{array}\right], \mathbf{b}_{2}=\left[\begin{array}{r}{1} \\ {-5}\end{array}\right], \mathbf{c}_{1}=\left[\begin{array}{l}{1} \\ {4}\end{array}\right], \mathbf{c}_{2}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right]\)

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 \(\left[\begin{array}{ccc}{1} & {-3} & {4} \\ {2} & {-6} & {8}\end{array}\right]=\mathbf{u v}^{T}\)

In Exercises 1鈥�4, assume that the matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find bases for Col A, Row A, and Nul A. $$ A=\left[\begin{array}{rrrrrr}{1} & {1} & {-3} & {7} & {9} & {-9} \\ {1} & {2} & {-4} & {10} & {13} & {-12} \\ {1} & {-1} & {-1} & {1} & {1} & {-3} \\ {1} & {-3} & {1} & {-5} & {-7} & {3} \\ {1} & {-2} & {0} & {0} & {-5} & {-4}\end{array}\right] $$ $$ B=\left[\begin{array}{rrrrrr}{1} & {1} & {-3} & {7} & {9} & {-9} \\ {0} & {1} & {-1} & {3} & {4} & {-3} \\ {0} & {0} & {0} & {1} & {-1} & {-2} \\ {0} & {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {0} & {0}\end{array}\right] $$

Let \(\mathcal{B}=\left\\{1, \cos t, \cos ^{2} t, \ldots, \cos ^{6} t\right\\}\) and \(\mathcal{C}=\\{1, \cos t\) \(\cos 2 t, \ldots, \cos 6 t \\} .\) Assume the following trigonometric identities (see Exercise 37 in Section 4.1 ). \(\cos 2 t=-1+2 \cos ^{2} t\) \(\cos 3 t=-3 \cos t+4 \cos ^{3} t\) \(\cos 4 t=1-8 \cos ^{2} t+8 \cos ^{4} t\) \(\cos 5 t=5 \cos t-20 \cos ^{3} t+16 \cos ^{5} t\) \(\cos 6 t=-1+18 \cos ^{2} t-48 \cos ^{4} t+32 \cos ^{6} t\) Let \(H\) be the subspace of functions spanned by the functions in \(\mathcal{B} .\) Then \(\mathcal{B}\) is a basis for \(H,\) by Exercise 38 in Section \(4.3 .\) a. Write the \(\mathcal{B}\) -coordinate vectors of the vectors in \(\mathcal{C},\) and use them to show that \(\mathcal{C}\) is a linearly independent set in \(H .\) b. Explain why \(\mathcal{C}\) is a basis for \(H\)
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