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Chapter 7: Problem 21

For the matrices \(A\) find \(\lim A^{t} .\) Feel free to use Theorem 7.4 .1 \(t \rightarrow \infty\). $$A=\left[\begin{array}{ll} 0.5 & 0.25 \\ 0.5 & 0.75 \end{array}\right]$$

Short Answer

Expert verified
Matrix A is not stochastic. To find the limit, compute high powers of A manually or with software and observe the pattern. However, without the explicit calculations or relevant theorem, we cannot provide the exact limit here.

Step by step solution

01

Understanding the Problem

We are asked to find the limit of the matrix powers of A as t tends to infinity, which is denoted by \(\lim_{t \to \infty} A^t\). To solve this, we will explore the behavior of the matrix A as it is raised to higher powers.
02

Verify if A is a Stochastic Matrix

By definition, a stochastic matrix has each row summing up to 1. We verify this for matrix A. The sum of each row in A is \(0.5 + 0.25 = 0.75\) and \(0.5 + 0.75 = 1.25\), which means A is not a stochastic matrix.
03

Finding the Eigenvalues

We calculate the eigenvalues of A by solving the characteristic equation \(\det(A - \lambda I) = 0\).
04

Diagonalizing A (If Possible)

If A has a set of linearly independent eigenvectors, it can be diagonalized as \(A = PDP^{-1}\), where P is the matrix of eigenvectors and D is the diagonal matrix of eigenvalues. Diagonalization can simplify the process of finding \(A^t\) as power calculation is straightforward with diagonal matrices.
05

Apply Theorem 7.4.1 if Relevant

Using Theorem 7.4.1, for a stochastic matrix A, with eigenvalues such that \(1 = \lambda_1 > |\lambda_2| \geq ... \geq |\lambda_n|\), it will hold that \(\lim_{t \to \infty} A^t\) equals a matrix with identical rows, each of which is a fixed distribution vector. For A which is not stochastic, we cannot directly apply Theorem 7.4.1.
06

Determining the Limit of A^t

We need to compute high powers of A manually or using software, and observe the pattern as t increases to see if the powers of A approach a specific matrix.

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Key Concepts

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

Stochastic Matrix
At the heart of many applications in economics, physics, and particularly in the field of Markov chains, lies the concept of a stochastic matrix. A stochastic matrix, also known as a probability matrix, is a square matrix used to describe transitions in a Markov chain. What makes a matrix stochastic is the fact that each of its rows represents a probability distribution: all the elements are non-negative, and the sum of the elements in each row is exactly one.

This requirement aligns with the idea that the total probability of moving from one state to another in a system must be 100%. However, not every matrix used in probability calculations is stochastic. As illustrated by the step-by-step solution where the matrix in question didn't meet the row sum criterion, it's essential to verify this property before applying probabilistic interpretations or certain theorems related to stochastic processes.
Eigenvalues
Eigenvalues are a fundamental part of understanding matrices in algebra. They are the special set of scalars associated with a system of linear equations that are derived from a square matrix. Comprehending eigenvalues is critical for various mathematical and physical problems, including those involving stability analysis and systems of differential equations.

In the matrix powers limit problem context, eigenvalues help to determine the long-term behavior of power iterations. This is because the eigenvalues of a matrix dictate how vectors are stretched or shrunk when they are multiplied by the matrix, influencing how the matrix behaves when raised to high powers. To find the eigenvalues, one must solve the characteristic equation which is a polynomial derived from the matrix. This calculation provides key insights into the matrix's properties.
Diagonalization
Diagonalization is a technique where a matrix is expressed in terms of its eigenvalues and eigenvectors. The process transforms a square matrix into a diagonal matrix, which preserves the eigenvalues but simplifies complex calculations, especially when it comes to raising the matrix to a large power. The diagonal matrix, often denoted by 'D', contains the eigenvalues on its diagonal, and the rest of the elements are zeroes.

When a matrix A is diagonalizable, it takes the form of A = PDP-1, where P is the matrix whose columns are the eigenvectors of A, and D is the diagonal matrix. This factorization is particularly useful because calculating powers of a diagonal matrix is straightforward; we just raise each diagonal entry (the eigenvalues) to the corresponding power. For matrices that are not diagonalizable, other methods need to be employed to understand their behavior under exponentiation.
Characteristic Equation
The characteristic equation is the equation we solve to find the eigenvalues of a matrix. Formally, it is derived from the matrix A by the equation \(\det(A - \lambda I) = 0\), where \(\lambda\) represents an eigenvalue and I is the identity matrix of the same size as A. The determinant of \(A - \lambda I\) gives us a polynomial where the roots are the eigenvalues.

Frequently, the characteristic equation is a foundational step in the process of diagonalization because it paves the way to finding both the eigenvalues and eigenvectors necessary for constructing the diagonal matrix D and the matrix of eigenvectors P. Understanding this equation is vital for students tackling linear algebra problems as it provides a gateway to a deeper comprehension of a matrix鈥檚 properties and behaviors.

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Most popular questions from this chapter

Three holy men (let's call them Anselm, Benjamin, and Caspar) put little stock in material things; their only earthly possession is a small purse with a bit of gold dust. Each day they get together for the following bizarre bonding ritual: Each of them takes his purse and gives his gold away to the two others, in equal parts. For example, if Anselm has 4 ounces one day, he will give 2 ounces each to Benjamin and Caspar. a. If Anselm starts out with 6 ounces, Benjamin with ounce, and Caspar with 2 ounces, find formulas for the amounts \(a(t), b(t),\) and \(c(t)\) each will have after distributions. Hint: The vectors \(\left[\begin{array}{l}1 \\ 1 \\\ 1\end{array}\right],\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right],\) and \(\left[\begin{array}{r}1 \\ 0 \\ -1\end{array}\right]\) will be useful. b. Who will have the most gold after one year, that is, after 365 distributions?

Consider the national income of a country, which consists of consumption, investment, and government expenditures. Here we assume the government expenditure to be constant, at \(G_{0},\) while the national income \(Y(t),\) consumption \(C(t),\) and investment \(I(t)\) change over time. According to a simple model, we have \\[ \begin{array}{l} Y(t)=C(t)+I(t)+G_{0} \\ C(t+1)=\gamma Y(t) \\ I(t+1)=\alpha(C(t+1)-C(t)) \end{array} | \begin{array}{l} (0<\gamma<1) \\ (\alpha>0) \end{array} \\] where \(\gamma\) is the marginal propensity to consume and \(\alpha\) is the acceleration coefficient. (See Paul E. Samuelson, "Interactions between the Multiplier Analysis and the Principle of Acceleration," Review of Economic Statistics, May \(1939,\) pp. \(75-78 .\) a. Find the equilibrium solution of these equations, when \(Y(t+1)=Y(t), C(t+1)=C(t),\) and \\[ I(t+1)=I(t) \\] b. Let \(y(t), c(t),\) and \(i(t)\) be the deviations of \(Y(t)\) \(C(t),\) and \(I(t),\) respectively, from the equilibrium state you found in part (a). These quantities are related by the equations \\[ \begin{array}{l} y(t)=c(t)+i(t) \\ c(t+1)=\gamma y(t) \\ i(t+1)=\alpha(c(t+1)-c(t)) \end{array} | \\] (Verify this!) By substituting \(y(t)\) into the second equation, set up equations of the form \\[ \left|\begin{array}{l} c(t+1)=p c(t)+q i(t) \\ i(t+1)=r c(t)+s i(t) \end{array}\right| \\] c. When \(\alpha=5\) and \(\gamma=0.2,\) determine the stability of the zero state of this system. d. When \(\alpha=1\) (and \(\gamma\) is arbitrary, \(0<\gamma<1\) ), determine the stability of the zero state. e. For each of the four sectors in the \(\alpha-\gamma\) -plane, determine the stability of the zero state. Discuss the various cases, in practical terms.

For the matrices \(A\) in Exercises 1 through \(10,\) determine whether the zero state is a stable equilibrium of the dynamical system \(\vec{x}(t+1)=A \vec{x}(t)\). $$A=\left[\begin{array}{rr} -1 & 3 \\ -1.2 & 2.6 \end{array}\right]$$

Consider the matrices A in Exercises 11 through \(16 .\) For which real numbers \(k\) is the zero state a stable equilibrium of the dynamical system \(\vec{x}(t+1)=A \vec{x}(t) ?\) $$A=\left[\begin{array}{rr} 0.7 & k \\ 0 & -0.9 \end{array}\right]$$

Leonardo of Pisa: The rabbit problem. Leonardo of Pisa (c. \(1170-1240\) ), also known as Fibonacci, was the first outstanding European mathematician after the ancient Greeks. He traveled widely in the Islamic world and studied Arabic mathematical writing. His work is in the spirit of the Arabic mathematics of his day. Fibonacci brought the decimal-position system to Europe. In his book Liber abaci \((1202),^{11}\) Fibonacci discusses the following problem: How many pairs of rabbits can be bred from one pair in one year? A man has one pair of rabbits at a certain place entirely surrounded by a wall. We wish to know how many pairs can be bred from it in one year, if the nature of these rabbits is such that they breed every month one other pair and begin to breed in the second month after their birth. Let the first pair breed a pair in the first month, then duplicate it and there will be 2 pairs in a month. From these pairs one, namely, the first, breeds a pair in the second month, and thus there are 3 pairs in the second month. From these, in one month, two will become pregnant, so that in the third month 2 pairs of rabbits will be born. Thus, there are 5 pairs in this month. From these, in the same month, 3 will be pregnant, so that in the fourth month there will be 8 pairs. From these pairs, 5 will breed 5 other pairs, which, added to the 8 pairs, gives 13 pairs in the fifth month, from which 5 pairs (which were bred in that same month) will not conceive in that month, but the other 8 will be pregnant. Thus, there will be 21 pairs in the sixth month. When we add to these the 13 pairs that are bred in the seventh month, then there will be in that month 34 pairs [and so on, \(55,89,144,233,377, \ldots .\). Finally, there will be 377 , and this number of pairs has been born from the first-mentioned pair at the given place in one year. Let \(j(t)\) be the number of juvenile pairs and \(a(t)\) the number of adult pairs after \(t\) months. Fibonacci starts his thought experiment in rabbit breeding with one adult pair, so \(j(0)=0\) and \(a(0)=1 .\) At \(t=1,\) the adult pair will have bred a (juvenile) pair, so \(a(1)=1\) and \(j(1)=1 .\) At \(t=2,\) the initial adult pair will have bred another (juvenile) pair, and last month's juvenile pair will have grown up, so \(a(2)=2\) and \(j(2)=1\) a. Find formulas expressing \(a(t+1)\) and \(j(t+1)\) in terms of \(a(t)\) and \(j(t) .\) Find the matrix \(A\) such that \\[ \vec{x}(t+1)=A \vec{x}(t) \\] where \\[ \vec{x}(t)=\left[\begin{array}{l} a(t) \\ j(t) \end{array}\right] \\]. b. Find closed formulas for \(a(t)\) and \(j(t) .\) (Note: You will have to deal with irrational quantities here. c. Find the limit of the ratio \(a(t) / j(t)\) as \(t\) approaches infinity. The result is known as the golden section. The golden section of a line segment \(A B\) is given by the point \(P\) such that \(\frac{\overline{A B}}{\overline{A P}}=\frac{\overline{A P}}{\overline{P B}}\).
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