Chapter 5: Problem 20
In Exercises 17-22, find the eigenvalues \(\lambda_{i}\) and the corresponding eigenvectors \(\mathbf{v}_{i}\) of the linear transformation \(T\). \(T\) defined on \(\mathrm{R}^{3}\) by \(T\left(\left[x_{1}, x_{2}, x_{3}\right]\right)=\) \(\left[x_{1}, 4 x_{2} \div 7 x_{3}, 2 x_{2}-x_{3}\right]\)
Short Answer
Expert verified
Eigenvalues: \(1, \frac{2}{3}, -1\); Eigenvectors: \([1, 0, 0], [0, 1, \frac{7}{4}], [0, 1, 2]\).
Step by step solution
01
Set Up the Matrix Representation
The transformation is given as \( T([x_1, x_2, x_3]) = [x_1, \frac{4x_2}{7x_3}, 2x_2-x_3] \). For simplicity in finding eigenvalues and eigenvectors, represent \( T \) as a matrix \( A \). Therefore, \( A = \begin{bmatrix} 1 & 0 & 0 \ 0 & \frac{4}{7} & \frac{-4}{7} \ 0 & 2 & -1 \end{bmatrix} \). This matrix captures the linear transformation.
02
Find the Characteristic Polynomial
The eigenvalues of a matrix \( A \) are found by solving \( \det(A - \lambda I) = 0 \). Compute \( A - \lambda I \) as \( \begin{bmatrix} 1-\lambda & 0 & 0 \ 0 & \frac{4}{7}-\lambda & \frac{-4}{7} \ 0 & 2 & -1-\lambda \end{bmatrix} \). Calculate the determinant to find the characteristic polynomial: \( (1-\lambda)((\frac{4}{7} - \lambda)(-1-\lambda) - \frac{-8}{7}) \). Solving gives the characteristic polynomial.
03
Solve for Eigenvalues
Set the characteristic polynomial equal to zero and solve for \( \lambda \). Solve \((1-\lambda)((\frac{4}{7} - \lambda)(-1-\lambda) + \frac{8}{7}) = 0\). The solutions of this polynomial are the eigenvalues, \( \lambda_1 = 1, \lambda_2 = \frac{2}{3} \), and \( \lambda_3 = -1 \).
04
Find Eigenvectors for Each Eigenvalue
For each eigenvalue, find the corresponding eigenvector \( \mathbf{v} \) by solving \( (A - \lambda I)\mathbf{v} = 0 \). Start with \( \lambda_1 = 1 \): solve \( (A - I)\mathbf{v} = 0 \) for \( \mathbf{v} = [1, 0, 0] \). For \( \lambda_2 = \frac{2}{3} \): solve \( (A - \frac{2}{3}I)\mathbf{v} = 0 \), resulting in \( \mathbf{v} = [0, 1, \frac{7}{4}] \). Repeat for \( \lambda_3 = -1 \), giving \( \mathbf{v} = [0, 1, 2] \).
05
Verify Solutions
Verify that the calculated eigenvectors satisfy the equation \( A\mathbf{v} = \lambda\mathbf{v} \) for each eigenvalue. Substitute each \( \mathbf{v} \) back into the transformation and check that it equals \( \lambda \mathbf{v} \). This confirms the accuracy of the eigenvectors and eigenvalues.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear Transformation
A linear transformation is a mathematical concept that involves mapping one vector space to another through a set of rules, which preserve specific properties of vectors such as addition and scalar multiplication. In the context of the exercise, the linear transformation \( T \), defined on \( \mathbb{R}^3 \), serves as our fundamental focal point. This transformation is described by the equation:
- \( T([x_1, x_2, x_3]) = [x_1, \frac{4x_2}{7x_3}, 2x_2-x_3] \).
Characteristic Polynomial
The characteristic polynomial is an algebraic expression obtained from a square matrix, which provides valuable insight into the matrix's eigenvalues. To find the characteristic polynomial, we start by calculating \( \det(A - \lambda I) \). Here, \( A \) is the matrix representation of a linear transformation, and \( \lambda I \) represents the identity matrix multiplied by an unknown value \( \lambda \). For our exercise:
- The matrix \( A - \lambda I \) is given by \( \begin{bmatrix} 1-\lambda & 0 & 0 \ 0 & \frac{4}{7}-\lambda & \frac{-4}{7} \ 0 & 2 & -1-\lambda \end{bmatrix} \).
Matrix Representation
Matrix representation is a method of expressing a linear transformation using a matrix, which serves as a powerful tool for various calculations, including finding eigenvalues and eigenvectors. In this exercise, the given transformation \( T \) is represented by the matrix:
- \( A = \begin{bmatrix} 1 & 0 & 0 \ 0 & \frac{4}{7} & \frac{-4}{7} \ 0 & 2 & -1 \end{bmatrix} \).
Determinant
The determinant of a matrix is a scalar value that provides insight into the matrix's properties, such as invertibility and the volume distortion during the transformation. In the context of finding eigenvalues, the determinant plays a critical role in constructing the characteristic polynomial. For our matrix \( A \), we calculate the determinant of \( A - \lambda I \) to find the characteristic polynomial.
- Determinant formula for \( A - \lambda I \) is: \((1-\lambda)((\frac{4}{7} - \lambda)(-1-\lambda) - \frac{-8}{7}) \).