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Chapter 10: Problem 7

Using the fact that the column sums of an exchange matrix \(E\) are all \(1,\) show that the column sums of \(I-E\) are zero. From this, show that \(I-E\) has zero determinant, and so \((I-E) \mathbf{p}=0\) has nontrivial solutions for \(\mathbf{p}\).

Short Answer

Expert verified
The column sums of \(I-E\) are zero, making \(I-E\) a singular matrix with zero determinant, leading to nontrivial solutions for \((I-E) \mathbf{p} = 0\).

Step by step solution

01

Understanding the exchange matrix

An exchange matrix \( E \) is a square matrix where each column contains exactly one \( 1 \) and all other entries are \( 0 \). This means for any column \( j \), \( \sum_{i} E_{ij} = 1 \). Since this holds for all columns, every column sum of \( E \) is \( 1 \).
02

Calculate column sums of \( I - E \)

The identity matrix \( I \) has a \( 1 \) in every diagonal entry. Hence, \((I-E)\) is created by subtracting \( E \) from \( I \), which fluctuates the diagonal entries to be \( 1 \) minus whatever entry is in \( E \) (typically \( 1 \)). Each column sum of \( (I-E) \) is \( \sum_i (I_{ij} - E_{ij}) = \sum_i I_{ij} - \sum_i E_{ij} = 1 - 1 = 0 \).
03

Show Zero Determinant

A matrix with zero column sums cannot be invertible since the columns are linearly dependent. Thus, \( I - E \) has a zero determinant because the sum of its columns nullifies the matrix.
04

Show Nontrivial Solutions Exist for \((I-E)\mathbf{p}=0\)

The determinant being zero implies that the matrix \( I-E \) is singular, meaning there exists a non-zero vector \( \mathbf{p} \) such that \((I-E)\mathbf{p} = 0\). This indicates the system of linear equations represented by \((I-E)\mathbf{p}=0\) has nontrivial solutions.

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Key Concepts

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

Linear Dependence
In the study of matrices, understanding linear dependence is fundamental. A set of vectors is said to be linearly dependent if one of the vectors can be expressed as a linear combination of the others. Applying this to columns of a matrix, if the columns are linearly dependent, it implies there exists a non-zero vector solution to the equation derived from multiplying the matrix and this vector.
This concept is crucial in solving linear systems because linear dependence indicates no unique solution or infinitely many solutions may exist for the system.
  • When \(I-E\) has column sums of zero, it implies the columns are linearly dependent.
  • This dependency ensures that multiplying \(I-E\) by a non-zero vector \(\mathbf{p}\) gives \((I-E) \mathbf{p}=0\), confirming nontrivial solutions.
Recognizing linear dependence allows for a deeper understanding of matrix behavior and is a stepping stone in understanding determinant properties.
Determinant
The determinant is a special number associated with square matrices that can reveal many characteristics about the matrix. The determinant can tell us if a matrix is invertible or not.
For the matrix \(I-E\), the fact that its determinant is zero is highly significant:
  • A zero determinant indicates the matrix is singular.
  • A singular matrix has linearly dependent columns and cannot be inverted.
This mathematical feature often underlines the existence of nontrivial solutions for the associated homogeneous equations. Calculating a determinant involves summing specific products of matrix entries, but ensuring that calculation yields zero showcases relationships among rows or columns.
Identity Matrix
The identity matrix, often denoted by \(I\), is pivotal in linear algebra.
It is an \(n \times n\) matrix that serves as the multiplicative identity in matrix algebra, meaning that \(I \mathbf{x} = \mathbf{x}\) for any vector \(\mathbf{x}\).
Key properties include:
  • Diagonally contains ones, with all other elements being zeros.
  • Acts as the 'do-nothing' matrix in multiplication, leaving matrices unchanged.
Regarding \(I-E\), the identity matrix \(I\) helps in expressing transformations and operations in a clearer manner. Subtracting an exchange matrix \(E\) from \(I\) alters specific entries while retaining overall structural identity properties, particularly impacting column sums, and emphasizing changes that lead to linear dependence and zero determinant findings.

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Most popular questions from this chapter

Suppose that a certain animal population is divided into two age classes and has a Leslie matrix $$ L=\left[\begin{array}{ll} 1 & \frac{3}{2} \\ \frac{1}{2} & 0 \end{array}\right] $$ (a) Calculate the positive eigenvalue \(\lambda_{1}\) of \(L\) and the corresponding eigenvector \(\mathbf{x}_{1}.\) (b) Beginning with the initial age distribution vector. $$\mathbf{x}^{(0)}=\left[\begin{array}{c} 100 \\ 0 \end{array}\right]$$ calculate \(x^{(1)}, x^{(2)}, x^{(3)}, x^{(4)},\) and \(x^{(S)},\) rounding off to the nearest integer when necessary. (c) Calculate \(x^{(6)}\) using the exact formula \(x^{(6)}=L x^{(S)}\) and using the approximation formula \(x^{(6)} \approx \lambda_{1} x^{(S)}.\) Answer: A. $$\lambda_{1}=\frac{3}{2}, \quad x_{1}=\left[\begin{array}{l} 1 \\ \frac{1}{3} \end{array}\right]$$ B. $$\mathbf{x}^{(1)}=\left[\begin{array}{c} 100 \\ 50 \end{array}\right], \mathbf{x}^{(2)}=\left[\begin{array}{c} 175 \\ 50 \end{array}\right], \mathbf{x}^{(3)}=\left[\begin{array}{c} 250 \\ 88 \end{array}\right], \mathbf{x}^{(4)}=\left[\begin{array}{l} 382 \\ 125 \end{array}\right], \mathbf{x}^{(5)}=\left[\begin{array}{l} 570 \\ 191 \end{array}\right]$$ C. $$\mathbf{x}^{(6)}=L \mathbf{x}^{(S)}=\left[\begin{array}{l} 857 \\ 285 \end{array}\right], \mathbf{x}^{(6)} \simeq \lambda_{1} \mathbf{x}^{(S)}=\left[\begin{array}{l} 855 \\ 287 \end{array}\right]$$

A Hill 2 -cipher is intercepted that starts with the pairs $$S L H K$$ Find the deciphering and enciphering matrices, given that the plaintext is known to start with the word \(A R M Y\).

(a) What can you say about the coefficients \(c_{1}, c_{2},\) and \(c_{3}\) that determine a convex combination \(\mathbf{v}=\mathbf{c}_{1} \mathbf{v}_{1}+\mathbf{c}_{2} \mathbf{v}_{2}+\mathbf{c}_{3} \mathbf{v}_{3}\) if \(v\) lies on one of the three vertices of the triangle determined by the three vectors v_, v_2, and v3? (b) What can you say about the coefficients \(c_{1}, c_{2},\) and \(c_{3}\) that determine a convex combination \(\mathbf{v}=\mathbf{c}_{1} \mathbf{v}_{1}+\mathbf{c}_{2} \mathbf{v}_{2}+\mathbf{c}_{3} \mathbf{v}_{3}\) if \(v\) lies on one of the three sides of the triangle determined by the three vectors \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3} ?\) (c) What can you say about the coefficients \(c_{1}, c_{2},\) and \(c_{3}\) that determine a convex combination \(\mathbf{v}=\mathbf{c}_{1} \mathbf{v}_{1}+\mathbf{c}_{2} \mathbf{v}_{2}+\mathbf{c}_{3} \mathbf{v}_{3}\) if \(v\) lies in the interior of the triangle determined by the three vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}\) and va?

A manufacturer produces sacks of chicken feed from two ingredicnts, \(A\) and \(B\). Each sack is to contain at least 10 ounces of nutrient \(N_{1}\), at least 8 ounces of nutrient \(N_{2}\), and at least 12 ounces of nutrient \(N_{3}\). Each pound of ingredicnt \(A\) contains 2 ounces of nutricnt \(N_{1}\), 2 ounces of nutricnt \(N_{2}\), and 6 ounces of nutrient \(N_{3}\). Each pound of ingredicnt \(B\) contains 5 ounces of nutricnt \(N_{1}\), 3 ounces of nutrient \(N_{2}\), and 4 ounces of nutrient \(N_{3} .\) If ingredicnt \(A\) costs 8 cents per pound and ingredicnt \(B\) costs 9 cents per pound, how much of each ingredient should the manufacturer use in each sack of feed to minimize his costs?

Find the equations of the planes in 3 -space that pass through the following points: (a) (1,1,-3),(1,-1,1),(0,-1,2) (b) \((2,3,1), \quad(2,-1,-1), \quad(1,2,1)\)
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