/*! 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 26 Exercises \(22-26\) provide a gl... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercises \(22-26\) provide a glimpse of some widely used matrix factorizations, some of which are discussed later in the text. (Spectral Factorization) Suppose a \(3 \times 3\) matrix A admits a factorization as \(A=P D P^{-1},\) where \(P\) is some invertible \(3 \times 3\) matrix and \(D\) is the diagonal matrix $$ D=\left[\begin{array}{ccc}{1} & {0} & {0} \\ {0} & {1 / 2} & {0} \\ {0} & {0} & {1 / 3}\end{array}\right] $$ Show that this factorization is useful when computing high powers of \(A .\) Find fairly simple formulas for \(A^{2}, A^{3},\) and \(A^{k}\) (k a positive integer), using \(P\) and the entries in \(D\) .

Short Answer

Expert verified
Use \( A^k = P D^k P^{-1} \) with \( D^k = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \frac{1}{2^k} & 0 \\ 0 & 0 & \frac{1}{3^k} \end{bmatrix} \) for matrix powers.

Step by step solution

01

Understanding the Matrix Factorization

The matrix \( A \) is expressed as \( A = P D P^{-1} \), where \( P \) is an invertible matrix, and \( D \) is a diagonal matrix.
02

Knowing Powers of Diagonal Matrix

For any power \( k \), the powers of a diagonal matrix \( D \) can be computed by raising each of its diagonal entries to the power \( k \). This yields \( D^k = \begin{bmatrix} 1^k & 0 & 0 \ 0 & (\frac{1}{2})^k & 0 \ 0 & 0 & (\frac{1}{3})^k \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & \frac{1}{2^k} & 0 \ 0 & 0 & \frac{1}{3^k} \end{bmatrix} \).
03

Calculating A^k using Spectral Factorization

Given the spectral factorization \( A = P D P^{-1} \), the k-th power of \( A \) can be represented as \( A^k = (P D P^{-1})^k = P D^k P^{-1} \). Since \( D^k \) is straightforward to compute, calculating \( A^k \) becomes significantly easier.
04

Calculating Specific Powers A^2 and A^3

Using the formula \( A^k = P D^k P^{-1} \):- For \( A^2 \), this becomes \( A^2 = P D^2 P^{-1} \) where \( D^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & \frac{1}{4} & 0 \ 0 & 0 & \frac{1}{9} \end{bmatrix} \).- For \( A^3 \), this becomes \( A^3 = P D^3 P^{-1} \) where \( D^3 = \begin{bmatrix} 1 & 0 & 0 \ 0 & \frac{1}{8} & 0 \ 0 & 0 & \frac{1}{27} \end{bmatrix} \).
05

Expressing General Formula for A^k

The general formula for high powers of \( A \) in the context of spectral factorization is \( A^k = P D^k P^{-1} \), where \( D^k = \begin{bmatrix} 1 & 0 & 0 \ 0 & \frac{1}{2^k} & 0 \ 0 & 0 & \frac{1}{3^k} \end{bmatrix} \). This method simplifies the process of computing high powers by taking advantage of the diagonal nature of \( D \).

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Key Concepts

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

Matrix Factorization
Matrix factorization is a powerful mathematical technique used to simplify complex problems, especially when dealing with large matrices. In the specific context of spectral factorization, we express a matrix \( A \) in the form \( A = P D P^{-1} \). Here, \( P \) is an invertible matrix that 'diagonalizes' \( A \), and \( D \) is a diagonal matrix.

The process of factorization transforms the matrix into a product of simpler matrices, making subsequent operations more manageable. Since \( D \) is a diagonal matrix, it is much simpler to manipulate compared to \( A \) directly. Understanding how to factorize matrices is essential for simplifying mathematical operations, like finding power operations or solving linear equations efficiently.

In practical terms, factorization is used not only in theoretical problems but also in fields like computer graphics, where transformations and manipulations of matrices are frequent tasks.
Diagonal Matrices
Diagonal matrices are matrices in which all the elements outside the main diagonal are zeros. Thus, a diagonal matrix \( D \) looks like \( \begin{bmatrix} d_1 & 0 & 0 \ 0 & d_2 & 0 \ 0 & 0 & d_3 \end{bmatrix} \), where each \( d_i \) are elements along the main diagonal.

The special property of diagonal matrices is that they are incredibly easy to work with in computations, particularly when it comes to matrix powers, inverses, and determinant calculations. For example, finding the inverse of a diagonal matrix simply requires taking the reciprocal of each diagonal element, assuming they are non-zero.

Additionally, the determinant of a diagonal matrix is the product of its diagonal entries \( (d_1 \cdot d_2 \cdot d_3) \), which simplifies calculations greatly. Because of these properties, diagonal matrices are fundamental in spectral factorization and linear algebra as a whole, making advanced computations more accessible.
Matrix Powers
Matrix powers involve multiplying a matrix by itself a given number of times. Calculating powers of matrices can often be complex, but when dealing with diagonal matrices, the task is significantly simplified.

For a diagonal matrix \( D \), raising it to the power of \( k \) means raising each diagonal element to that power: \( D^k = \begin{bmatrix} d_1^k & 0 & 0 \ 0 & d_2^k & 0 \ 0 & 0 & d_3^k \end{bmatrix} \). This is because the off-diagonal elements remain zero, no matter the power.

In spectral factorization, this simplification allows us to compute high powers of a matrix \( A \) efficiently, by using the relation \( A^k = P D^k P^{-1} \). Thus, handling matrix powers isn't just about repeated multiplication; it's about transforming the problem into a form where the multiplication becomes trivial, thanks to the properties of diagonal matrices.

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Most popular questions from this chapter

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations. [M] Describe in words what happens when you compute \(A^{5}\) , \(A^{10}, A^{20},\) and \(A^{30}\) for $$ A=\left[\begin{array}{ccc}{1 / 6} & {1 / 2} & {1 / 3} \\ {1 / 2} & {1 / 4} & {1 / 4} \\ {1 / 3} & {1 / 4} & {5 / 12}\end{array}\right] $$

Consider the production model \(\mathbf{x}=C \mathbf{x}+\mathbf{d}\) for an economy with two sectors, where $$C=\left[\begin{array}{cc}{.0} & {.5} \\ {.6} & {.2}\end{array}\right], \quad \mathbf{d}=\left[\begin{array}{c}{50} \\ {30}\end{array}\right]$$ Use an inverse matrix to determine the production level necessary to satisfy the final demand.

Exercises 1–4 refer to an economy that is divided into three sectors—manufacturing, agriculture, and services. For each unit of output, manufacturing requires .10 unit from other companies in that sector, .30 unit from agriculture, and .30 unit from services. For each unit of output, agriculture uses .20 unit of its own output, .60 unit from manufacturing, and .10 unit from services. For each unit of output, the services sector consumes .10 unit from services, .60 unit from manufacturing, but no agricultural products. Determine the production levels needed to satisfy a final demand of 18 units for manufacturing, 18 units for agriculture, and 0 units for services.

The Leontief production equation, \(\mathbf{x}=C \mathbf{x}+\mathbf{d},\) is usually accompanied by a dual price equation, \(\mathbf{p}=C^{T} \mathbf{p}+\mathbf{v}\) where \(\mathbf{p}\) is a price vector whose entries list the price per unit for each sector's output, and \(\mathbf{v}\) is a value added vector whose entries list the value added per unit of output. Value added includes wages, profit, depreciation, etc. ) An important fact in economics is that the gross domestic product (GDP) can be expressed in two ways: \(\\{\text { gross domestic product }\\}=\mathbf{p}^{T} \mathbf{d}=\mathbf{v}^{T} \mathbf{x}\) Verify the second equality. [Hint: Compute \(\mathbf{p}^{T} \mathbf{x}\) in two ways.

Construct a nonzero \(3 \times 3\) matrix \(A\) and a vector \(\mathbf{b}\) such that \(\mathbf{b}\) is not in \(\operatorname{Col} A\)

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