Chapter 8: Problem 23
In Problems, find the eigenvalues and eigenvectors of the given nonsingular matrix \(\mathbf{A} .\) Then without finding \(\mathbf{A}^{-1}\), find its eigenvalues and corresponding eigenvectors. $$ \mathbf{A}=\left(\begin{array}{ll} 5 & 1 \\ 1 & 5 \end{array}\right) $$
Short Answer
Expert verified
Eigenvalues: \(6, 4\). Eigenvectors: \((1, 1)\), \((1, -1)\). For \(\mathbf{A}^{-1}\), eigenvalues \(\frac{1}{6}, \frac{1}{4}\); eigenvectors same.
Step by step solution
01
Find the Characteristic Polynomial
To find the eigenvalues, we first find the characteristic polynomial of the matrix \( \mathbf{A} \). The characteristic polynomial is given by the determinant \( \text{det}(\mathbf{A} - \lambda \mathbf{I}) \). Here, \( \mathbf{I} \) is the identity matrix of the same size as \( \mathbf{A} \), and \( \lambda \) represents the eigenvalues.Therefore, we calculate:\(\text{det}(\begin{pmatrix} 5-\lambda & 1 \ 1 & 5-\lambda \end{pmatrix}) = (5-\lambda)(5-\lambda) - 1 \times 1\)This simplifies to:\((5-\lambda)^2 - 1 = \lambda^2 - 10\lambda + 24\)So the characteristic polynomial is \( \lambda^2 - 10\lambda + 24 \).
02
Solve the Characteristic Polynomial
Next, we solve the characteristic polynomial \( \lambda^2 - 10\lambda + 24 = 0 \) to find the eigenvalues.We factor the quadratic equation:\((\lambda - 6)(\lambda - 4) = 0\)Thus, the eigenvalues are \( \lambda_1 = 6 \) and \( \lambda_2 = 4 \).
03
Find Eigenvectors for Eigenvalue 6
Substitute \( \lambda_1 = 6 \) into \( \mathbf{A} - \lambda \mathbf{I} \) and set the resulting matrix equation to zero to find the eigenvector:\(\begin{pmatrix} -1 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\)Solving the system of equations, we observe:1. \( -x + y = 0 \)2. \( x - y = 0 \)From either equation, we find that \( y = x \). Thus, the eigenvector for \( \lambda_1 = 6 \) can be written as \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
04
Find Eigenvectors for Eigenvalue 4
Now, substitute \( \lambda_2 = 4 \) into \( \mathbf{A} - \lambda \mathbf{I} \) and solve the resulting matrix equation:\(\begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\)Solving the system of equations, we have:1. \( x + y = 0 \)2. \( x + y = 0 \)From these, \( y = -x \). Thus, the eigenvector for \( \lambda_2 = 4 \) can be represented as \( \begin{pmatrix} 1 \ -1 \end{pmatrix} \).
05
Find Eigenvalues and Eigenvectors of A^{-1}
The eigenvalues of \( \mathbf{A}^{-1} \) are the reciprocals of the eigenvalues of \( \mathbf{A} \). Therefore, for eigenvalues \( 6 \) and \( 4 \) of \( \mathbf{A} \), the eigenvalues of \( \mathbf{A}^{-1} \) are \( \frac{1}{6} \) and \( \frac{1}{4} \).The eigenvectors remain the same: \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \) for \( \frac{1}{6} \) and \( \begin{pmatrix} 1 \ -1 \end{pmatrix} \) for \( \frac{1}{4} \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Polynomial
The characteristic polynomial is a crucial part of finding eigenvalues of a square matrix. For a given matrix \(\mathbf{A}\), the characteristic polynomial is derived from the determinant of \(\mathbf{A} - \lambda \mathbf{I}\), where \(\lambda\) are the eigenvalues and \(\mathbf{I}\) is the identity matrix of the same dimension as \(\mathbf{A}\). This determinant is a polynomial expression in terms of \(\lambda\).
For our matrix \(\mathbf{A} = \begin{pmatrix} 5 & 1 \ 1 & 5 \end{pmatrix}\), the expression becomes:
This polynomial is what we solve to uncover the eigenvalues of \(\mathbf{A}\). Understanding how to derive this polynomial is a key skill in linear algebra and offers insights into the nature of the matrix.
For our matrix \(\mathbf{A} = \begin{pmatrix} 5 & 1 \ 1 & 5 \end{pmatrix}\), the expression becomes:
- \[\text{det}\begin{pmatrix} 5-\lambda & 1 \ 1 & 5-\lambda \end{pmatrix} = (5-\lambda)^2 - 1 \]
This polynomial is what we solve to uncover the eigenvalues of \(\mathbf{A}\). Understanding how to derive this polynomial is a key skill in linear algebra and offers insights into the nature of the matrix.
Matrix Inversion
Inverting a matrix is a method that provides significant insights into its properties, including eigenvalues and eigenvectors. The inverse of a matrix \(\mathbf{A}^{-1}\) is a matrix that when multiplied by \(\mathbf{A}\), results in the identity matrix \(\mathbf{I}\).
Although we did not directly compute \(\mathbf{A}^{-1}\) in this exercise, understanding inversion helps in exploring eigenvalues of \(\mathbf{A}^{-1}\) itself. For a nonsingular matrix like \(\mathbf{A}\), its inverse exists, and the eigenvalues of the inverse are simply the reciprocals of the original eigenvalues.
This concept highlights a fascinating relation: if \(\lambda_1, \lambda_2, \ldots\) are eigenvalues of \(\mathbf{A}\), then \(1/\lambda_1, 1/\lambda_2, \ldots\) are those of \(\mathbf{A}^{-1}\). This reciprocal relationship is particularly useful for understanding behaviors in different transformations and solving linear equations.
Although we did not directly compute \(\mathbf{A}^{-1}\) in this exercise, understanding inversion helps in exploring eigenvalues of \(\mathbf{A}^{-1}\) itself. For a nonsingular matrix like \(\mathbf{A}\), its inverse exists, and the eigenvalues of the inverse are simply the reciprocals of the original eigenvalues.
This concept highlights a fascinating relation: if \(\lambda_1, \lambda_2, \ldots\) are eigenvalues of \(\mathbf{A}\), then \(1/\lambda_1, 1/\lambda_2, \ldots\) are those of \(\mathbf{A}^{-1}\). This reciprocal relationship is particularly useful for understanding behaviors in different transformations and solving linear equations.
Quadratic Equation Factorization
Solving the characteristic polynomial often involves factoring a quadratic equation, which is typically represented as \(ax^2 + bx + c = 0\). This task is fundamental in finding the roots that correspond to the eigenvalues of a matrix.
In our example, the polynomial \(\lambda^2 - 10\lambda + 24\) factors into \((\lambda - 6)(\lambda - 4) = 0\). These factors directly reveal the eigenvalues: \(\lambda_1 = 6\) and \(\lambda_2 = 4\).
In our example, the polynomial \(\lambda^2 - 10\lambda + 24\) factors into \((\lambda - 6)(\lambda - 4) = 0\). These factors directly reveal the eigenvalues: \(\lambda_1 = 6\) and \(\lambda_2 = 4\).
- Identifying the coefficients in the polynomial (here \(a=1\), \(b=-10\), \(c=24\)) can streamline the process.
- Using factorization techniques like grouping or the quadratic formula helps tackle this task effectively.
Determinant Calculation
The determinant is a vital tool in linear algebra, particularly in solving systems of linear equations, determining matrix invertibility, and finding eigenvalues. It is a scalar quantity that provides insights into the matrix's characteristics.
For a 2x2 matrix \(\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as \(ad-bc\).
In the context of the characteristic polynomial for a matrix \(\mathbf{A}\), you compute the determinant of \(\mathbf{A} - \lambda \mathbf{I}\) to obtain the polynomial. Understanding and calculating determinants efficiently shows how matrices behave under various operations.
For a 2x2 matrix \(\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}\), the determinant is calculated as \(ad-bc\).
In the context of the characteristic polynomial for a matrix \(\mathbf{A}\), you compute the determinant of \(\mathbf{A} - \lambda \mathbf{I}\) to obtain the polynomial. Understanding and calculating determinants efficiently shows how matrices behave under various operations.
- Determinants help confirm if a square matrix is invertible. A non-zero determinant indicates that the matrix is invertible.
- They serve as a gateway to further understand systemic behaviors in mathematical transformations.