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Let \(A=\left[\begin{array}{ll}\lambda & 1 \\ 0 & \lambda\end{array}\right]\), and let \(E=\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right] .\) Use mathematical induction or the binomial formula to show that \(A^{m}=\lambda^{m} I+m \lambda^{m-1} E\).

Short Answer

Expert verified
Question: Prove that for a given matrix \(A = \begin{bmatrix}\lambda & 1 \\ 0 & \lambda \end{bmatrix}\), where λ is a scalar constant, the formula \(A^m = \lambda^m I + m \lambda^{m-1} E\) holds for all positive integer values of m using mathematical induction. Answer: By mathematical induction, we have shown that the formula \(A^m = \lambda^m I + m \lambda^{m-1} E\) holds for all positive integer values of m.

Step by step solution

01

Base Case (m = 1)

We need to check if the formula holds when m = 1. In this case, the formula should be \(A^1 = \lambda^1 I + 1 \lambda^{1-1} E\). We compute the value for \(A^1\): \(A^1 = A = \begin{bmatrix}\lambda & 1 \\ 0 & \lambda \end{bmatrix}\) Now, we compute the right-hand side of the formula: \(\lambda^1 I + 1 \lambda^{1-1} E = \begin{bmatrix}\lambda & 0 \\ 0 & \lambda \end{bmatrix} + \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix}\lambda & 1 \\ 0 & \lambda \end{bmatrix}\) Since both sides are equal, the formula holds for m = 1.
02

Inductive Step

Now, let's assume that the formula works for m = k, i.e., \(A^k = \lambda^k I + k \lambda^{k-1} E\). (1) We aim to prove that the formula also works for m = k + 1: \(A^{k+1} = \lambda^{k+1} I + (k+1) \lambda^k E\). (2) We start calculating the left-hand side of the formula (2) using the fact that \(A^{k+1} = A^k \cdot A\): \(A^{k+1} = (\lambda^k I + k \lambda^{k-1} E) \cdot A\). Now, we need to multiply the two matrices: \((\lambda^k I + k \lambda^{k-1} E) \cdot A = \left(\begin{bmatrix}\lambda^k & 0 \\ 0 & \lambda^k \end{bmatrix} + \begin{bmatrix}0 & k\lambda^{k-1} \\ 0 & 0 \end{bmatrix}\right) \cdot \begin{bmatrix}\lambda & 1 \\ 0 & \lambda \end{bmatrix}\). \(= \left(\begin{bmatrix}\lambda^k & k\lambda^{k-1} \\ 0 & \lambda^k \end{bmatrix}\right) \cdot \begin{bmatrix}\lambda & 1 \\ 0 & \lambda \end{bmatrix}\). \(= \begin{bmatrix}\lambda^{k+1} & (k+1)\lambda^k \\ 0 & \lambda^{k+1} \end{bmatrix}\). By comparing this result to the right-hand side of the formula (2), we can see that for m = k + 1: \(A^{k+1} = \lambda^{k+1} I + (k+1) \lambda^k E\). As a result, we have proved that if the formula works for m = k, it also works for m = k + 1.
03

Conclusion

By mathematical induction, we have shown that the formula \(A^m = \lambda^m I + m \lambda^{m-1} E\) holds for all positive integer values of m.

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Key Concepts

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

Matrix Exponentiation
Matrix exponentiation is the process of raising a matrix to a certain power, much like raising a number to a power. In linear algebra, calculating powers of matrices is a common task, especially in solving systems of linear equations or analyzing linear transformations. For a square matrix \( A \), the power \( A^m \) means multiplying the matrix \( A \) by itself \( m \) times. In mathematical induction problems, it's essential to establish a base case and then prove that the pattern holds for all subsequent cases. This often involves examining the matrix product calculations and simplifying the results using matrix arithmetic rules.

Matrix exponentiation becomes particularly relevant when working with transition matrices in Markov chains or in computing discrete dynamical systems where the repeated application of a transformation is studied. Understanding how these powers behave can simplify complex calculations and provide deeper insights into the properties of the system being modeled.
Binomial Formula
The binomial formula is a powerful tool in algebra that expresses the expansion of powers of binomials. When applied to matrices, it helps in understanding how to distribute powers across terms that follow binomial expressions. The binomial formula is given by:
  • \( (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \)
Although this expansion was originally defined for scalars, it can be creatively applied in matrix operations if the matrices involved commute (i.e., \( AB = BA \)).

In the context of this exercise, the binomial formula metaphorically helps in structuring the proof, as the expression \( A^m = \lambda^m I + m \lambda^{m-1} E \) shares a similar layered structure to that of binomial expansions. Each matrix in this expansion needs to be carefully constructed and understood, ensuring the formulations align with the properties of matrix multiplication.
Inductive Proof
Inductive proof is a powerful logical tool used to validate mathematical statements regarding natural numbers. The idea is to prove a base case (usually the simplest case, \( m = 1 \)), and then show that if a statement holds for an arbitrary case \( m = k \), it must also hold for \( m = k + 1 \). This type of proof is considered complete once both steps are successfully demonstrated, assuming the statement is true for all positive integers.

In this exercise, the base case involves calculating \( A^1 \) and showing it matches the form \( \lambda^1 I + 1 \lambda^{1-1} E \). The inductive step involves assuming that the formula works for \( m = k \), then proving it's true for \( m = k + 1 \). This showcases the recursive nature of the problem solution, where each step builds on the previous one to establish the overall validity of the equation for any integer \( m \).

Inductive proofs like this one utilize the concept of mathematical induction not only to solve problems but to reinforce understanding of how mathematical properties persist across different domains. It's a crucial technique in ensuring that theoretical results hold universally, creating a basis for more advanced mathematical exploration.
Linear Algebra
Linear algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. It involves studying lines, planes, and subspaces but extends to transformations that preserve vector addition and scalar multiplication. Linear algebra is fundamental in various applications such as computer graphics, engineering solutions, and more.

Matrix operations constitute a significant part of linear algebra. Understanding matrices allows the representation and manipulation of linear mappings. They are not mere grids of numbers; they represent rich mathematical structures used for computation and transformations. In particular, calculating matrix powers and analyzing their expressions can provide insights into eigenvalues and eigenvectors, which are critical for understanding the geometry of linear transformations.

This exercise shows how linear algebra concepts intertwine with others like mathematical induction to prove statements. The matrices \( A \) and \( E \) symbolize linear operators, and exponents of \( A \) model repeated applications of these operators. Therefore, linear algebra provides the framework through which the problem can be both intuitively and formally understood.

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Most popular questions from this chapter

Each of the systems of linear differential equations can be expressed in the form \(\mathbf{y}^{\prime}=P(t) \mathbf{y}+\mathbf{g}(t) .\) Determine \(P(t)\) and \(\mathbf{g}(t)\) $$ A^{\prime}(t)=\left[\begin{array}{cc} t^{-1} & 4 t \\ 5 & 3 t^{2} \end{array}\right], \quad A(1)=\left[\begin{array}{rr} 2 & 5 \\ 1 & -2 \end{array}\right] $$

Consider the differential equation \(\mathbf{y}^{\prime}=\left[\begin{array}{ll}2 & 1 \\ 0 & 2\end{array}\right] \mathbf{y}\). Example 2 shows that the corresponding exponential matrix is \(e^{A t}=\left[\begin{array}{cc}e^{2 t} & t e^{2 t} \\ 0 & e^{2 t}\end{array}\right] .\) Suppose that \(\mathbf{y}(1)=\left[\begin{array}{l}1 \\\ 2\end{array}\right] .\) Use the propagator property \((8)\) to determine \(\mathbf{y}(4)\) and \(\mathbf{y}(-1)\).

Each initial value problem was obtained from an initial value problem for a higher order scalar differential equation. What is the corresponding scalar initial value problem? $$ \mathbf{y}^{\prime}=\left[\begin{array}{c} y_{2} \\ y_{3} \\ y_{4} \\ y_{2}+y_{3} \sin \left(y_{1}\right)+y_{3}^{2} \end{array}\right], \quad \mathbf{y}(1)=\left[\begin{array}{r} 0 \\ 0 \\ -1 \\ 2 \end{array}\right] $$

Consider the \(R L\) network shown in the figure. Assume that the loop currents \(I_{1}\) and \(I_{2}\) are zero until a voltage source \(V_{S}(t)\), having the polarity shown, is turned on at time \(t=0 .\) Applying Kirchhoff's voltage law to each loop, we obtain the equations $$ \begin{aligned} -V_{S}(t)+L_{1} \frac{d I_{1}}{d t}+R_{1} I_{1}+R_{3}\left(I_{1}-I_{2}\right) &=0 \\ R_{3}\left(I_{2}-I_{1}\right)+R_{2} I_{2}+L_{2} \frac{d I_{2}}{d t} &=0 \end{aligned} $$ (a) Formulate the initial value problem for the loop currents, \(\left[\begin{array}{l}I_{1}(t) \\ I_{2}(t)\end{array}\right]\), assuming that $$ L_{1}=L_{2}=0.5 H, \quad R_{1}=R_{2}=1 k \Omega, \quad \text { and } \quad R_{3}=2 k \Omega . $$ (b) Determine a fundamental matrix for the associated linear homogeneous system. (c) Use the method of variation of parameters to solve the initial value problem for the case where \(V_{S}(t)=1\) for \(t>0\).

Suppose the Runge-Kutta method (12) is applied to the initial value problem \(\mathbf{y}^{\prime}=\) \(A \mathbf{y}, \mathbf{y}(0)=\mathbf{y}_{0}\), where \(A\) is a constant square matrix [thus, \(\left.\mathbf{f}(t, \mathbf{y})=A \mathbf{y}\right] .\) (a) Express each of the vectors \(\mathbf{K}_{j}\) in terms of \(h, A\), and \(\mathbf{y}_{k}, j=1,2,3,4\). (b) Show that the Runge-Kutta method, when applied to this initial value problem, can be unraveled to obtain $$ \mathbf{y}_{k+1}=\left(I+h A+\frac{h^{2}}{2 !} A^{2}+\frac{h^{3}}{3 !} A^{3}+\frac{h^{4}}{4 !} A^{4}\right) \mathbf{y}_{k} $$ (c) Use the differential equation \(\mathbf{y}^{\prime}=A \mathbf{y}\) to express the \(n\)th derivative, \(\mathbf{y}^{(n)}(t)\), in terms of \(A\) and \(\mathbf{y}(t)\). Express the Taylor series expansion $$ \mathbf{y}(t+h)=\sum_{n=0}^{\infty} \mathbf{y}^{(n)}(t) \frac{h^{n}}{n !} $$ in terms of \(h, A\), and \(\mathbf{y}(t)\). Compare the Taylor series with the right-hand side of (15), with \(t=t_{k}\) and \(\mathbf{y}\left(t_{k}\right)=\mathbf{y}_{k}\). How well does (15) replicate the Taylor series?

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