/*! 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 31 Let \(A\) be an \((n \times n)\)... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 31

Let \(A\) be an \((n \times n)\) matrix with eigenvalue \(\lambda\) and eigenvector \(\mathbf{x}\). Let \(\alpha\) be any constant. Use the definition, \(A \mathbf{x}=\lambda \mathbf{x}, \mathbf{x} \neq \mathbf{0}\), to show that \(\lambda+\alpha\) is an eigenvalue of the matrix \(A+\alpha I\) and that \(\alpha \lambda\) is an eigenvalue of \(\alpha A\). Similarly, if \(A\) is invertible, show that \(\lambda \neq 0\) and that \(1 / \lambda\) is an eigenvalue of \(A^{-1}\).

Short Answer

Expert verified
Based on the step by step solution, what we've shown is: 1. If 位 is an eigenvalue of matrix A with eigenvector x, then 位+伪 is an eigenvalue of (A+伪I) with the same eigenvector x. 2. If 位 is an eigenvalue of matrix A with eigenvector x, then 伪位 is an eigenvalue of (伪A) with the same eigenvector x. 3. If A is invertible, its eigenvalues cannot be equal to zero. 4. If 位 is an eigenvalue of an invertible matrix A with eigenvector x, then 1/位 is an eigenvalue of A鈦宦� with the same eigenvector x. In summary, the relationship between eigenvalues and eigenvectors of a given matrix A and its transformations, such as A + 伪I and 伪A, has been demonstrated. Additionally, we have shown the relationship between the eigenvalues of A and its inverse, if A is invertible.

Step by step solution

01

Show that 位+伪 is an eigenvalue of A+伪I

First, we'll demonstrate that 位+伪 is the eigenvalue of the matrix (A+伪I). Start with the equation \(A\mathbf{x} = \lambda\mathbf{x}\), and apply the following operations: \((A+\alpha I)\mathbf{x} = A\mathbf{x} + \alpha (I\mathbf{x}) = \lambda\mathbf{x} + \alpha\mathbf{x}\) As \(\alpha (I\mathbf{x}) = \alpha\mathbf{x}\), the equation simplifies to: \((A+\alpha I)\mathbf{x} = (\lambda+\alpha)\mathbf{x}\) Thus, the eigenvalue of the matrix (A+伪I) is 位+伪, with the same eigenvector.
02

Show that 伪位 is an eigenvalue of 伪A

Next, we will show that 伪位 is the eigenvalue of the matrix 伪A. Start with the equation \(A\mathbf{x} = \lambda\mathbf{x}\), and apply the following operations: \((\alpha A)\mathbf{x} = \alpha (A\mathbf{x}) = \alpha (\lambda\mathbf{x})\) As \(\alpha (\lambda\mathbf{x}) = (\alpha\lambda)\mathbf{x}\), the equation simplifies to: \((\alpha A)\mathbf{x} = (\alpha\lambda)\mathbf{x}\) Thus, the eigenvalue of the matrix 伪A is 伪位, with the same eigenvector.
03

Show that 位 鈮� 0 if A is invertible

If A is invertible, then the equation \(A\mathbf{x} = \lambda\mathbf{x}\) can be rewritten as: \(\lambda\mathbf{x} = A\mathbf{x} \Rightarrow \lambda I\mathbf{x} = A\mathbf{x} \Rightarrow (\lambda I - A)\mathbf{x} = \mathbf{0}\) If 位 = 0, then the equation turns into -A\(\mathbf{x}\) = \(\mathbf{0}\). But since A is invertible, it has no non-zero eigenvector corresponding to the zero eigenvalue. Hence, 位 鈮� 0 when A is invertible.
04

Show that 1/位 is an eigenvalue of A鈦宦�

Finally, we'll show that if A is invertible, 1/位 is an eigenvalue of A鈦宦�. Start with the equation \(A\mathbf{x} = \lambda\mathbf{x}\), and multiply both sides with the inverse of A: \(A^{-1}(A\mathbf{x}) = A^{-1}(\lambda\mathbf{x})\) As \(A^{-1}(A\mathbf{x}) = \mathbf{x}\) and \(A^{-1}(\lambda\mathbf{x}) = \lambda^{-1}(A^{-1}\mathbf{x})\), the equation simplifies to: \(A^{-1}\mathbf{x} = (\lambda^{-1})\mathbf{x}\) Thus, the eigenvalue of A鈦宦� is 1/位, with the same eigenvector.

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Key Concepts

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

Matrix Algebra
Matrix algebra is a cornerstone of linear algebra, and it provides a framework for both theoretical and practical applications in various fields, including physics, engineering, and economics. A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns.

The algebraic operations of matrices, such as addition, subtraction, multiplication, and scalar multiplication, follow specific rules. For instance, matrix addition requires two matrices of the same dimension, where corresponding elements are added together. Similarly, matrix multiplication involves the dot product of rows and columns, and it's important to note that matrix multiplication is not commutative, meaning that the order in which matrices are multiplied matters.

In the context of eigenvalues and eigenvectors, matrix algebra helps to articulate the relationship between a matrix and its linear transformation. The equation A\mathbf{x} = \lambda\mathbf{x} represents an eigenvector \(\mathbf{x}\) corresponding to an eigenvalue \(\lambda\) for the matrix \(A\). This relationship showcases the power of matrix algebra in simplifying complex systems and provides a way to diagnose the behavior of linear transformations.
Linear Transformations
Linear transformations are mappings between vector spaces that preserve the operations of vector addition and scalar multiplication. In essence, a linear transformation is a function, often represented by a matrix, that acts on vectors to produce new vectors in the same space or in a different space.

The study of eigenvalues and eigenvectors arises naturally within the theory of linear transformations. An eigenvector is a vector that is only scaled, not rotated or otherwise altered, by a linear transformation, and the eigenvalue is the factor by which it is scaled. The relationship A\mathbf{x} = \lambda\mathbf{x} indicates that applying the linear transformation represented by matrix \(A\) to eigenvector \(\mathbf{x}\) results in a vector parallel to \(\mathbf{x}\), scaled by the eigenvalue \(\lambda\).

Understanding eigenvectors and eigenvalues offers valuable insights into the nature of linear transformations, such as identifying directions of invariant action and understanding the structure of the transformation.
Inverse Matrices
An inverse matrix, denoted as \(A^{-1}\), is a matrix that, when multiplied with the original matrix \(A\), yields the identity matrix \(I\), effectively 'undoing' the action of \(A\). However, not every matrix has an inverse; only those that are square (having the same number of rows and columns) and non-singular (having a non-zero determinant) possess an inverse.

In the context of eigenvalues and eigenvectors, if a matrix \(A\) is invertible and has eigenvalue \(\lambda\), then the inverse matrix \(A^{-1}\) has an eigenvalue of \(\lambda^{-1}\). This is because applying \(A^{-1}\) essentially reverses the scaling effect imposed by \(A\) on its eigenvectors. Thus, the original operation of scaling an eigenvector by the eigenvalue \(\lambda\) is undone by the inverse operation, which scales by \(\lambda^{-1}\). This property helps us understand the deep connection between a matrix, its inverse, and the corresponding eigenvalues and eigenvectors.

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Most popular questions from this chapter

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 \(t>0\).

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 \(\mathbf{g}(t)\) $$ \begin{aligned} &y_{1}^{\prime}=t^{2} y_{1}+3 y_{2}+\sec t \\ &y_{2}^{\prime}=(\sin t) y_{1}+t y_{2}-5 \end{aligned} $$

In each exercise, find the general solution of the homogeneous linear system and then solve the given initial value problem. $$ \mathbf{y}^{\prime}=\left[\begin{array}{lll} 3 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{array}\right] \mathbf{y}, \quad \mathbf{y}(0)=\left[\begin{array}{r} -1 \\ 1 \\ 6 \end{array}\right] \quad \begin{array}{r} \text { [For Exercise 7, the characteristic } \\ \text { polynomial is } \left.p(\lambda)=-(\lambda-5)(\lambda-2)^{2} .\right] \end{array} $$

Give an example that shows that while similar matrices have the same eigenvalues, they may not have the same eigenvectors.

For the problem in the exercise specified, (a) Write a program that carries out Euler's method. Use a step size of \(h=0.01\). (b) Run your program on the interval given.(c) Check your numerical solution by comparing the first two values, \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\), with the hand calculations. (d) Plot the components of the numerical solution on a common graph over the time interval of interest.Exercise 4
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