Chapter 5: Problem 11
In Exercises 2-16, find the characteristic polynomial, the real eigenvalues, and the corresponding eigenvectors of the given matrix. $$ \left[\begin{array}{rrr} -2 & 0 & 0 \\ -5 & -2 & -5 \\ 5 & 0 & 3 \end{array}\right] $$
Short Answer
Expert verified
The eigenvalues are \(-2\) (multiplicity 2) and \(3\); corresponding eigenvectors are \(\begin{pmatrix} x_1 \\ 0 \\ -x_1 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\).
Step by step solution
01
Define the Matrix and Characteristic Polynomial Equation
We begin with the matrix \( A = \begin{pmatrix} -2 & 0 & 0 \ -5 & -2 & -5 \ 5 & 0 & 3 \end{pmatrix} \). The characteristic polynomial of a matrix is given by the determinant of \( A - \lambda I \), where \( \lambda \) is a scalar (the eigenvalue) and \( I \) is the identity matrix of the same size.
02
Subtract \( \lambda I \) from Matrix \( A \)
The identity matrix \( I \) is \( \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \). Thus, \( \lambda I = \begin{pmatrix} \lambda & 0 & 0 \ 0 & \lambda & 0 \ 0 & 0 & \lambda \end{pmatrix} \).Then, \( A - \lambda I = \begin{pmatrix} -2-\lambda & 0 & 0 \ -5 & -2-\lambda & -5 \ 5 & 0 & 3-\lambda \end{pmatrix} \).
03
Calculate the Determinant
The determinant of \( A - \lambda I \) is calculated by expanding along the first row (since it has zeros which simplify the computation):\[ \det(A - \lambda I) = (-2-\lambda) \left| \begin{array}{cc} -2-\lambda & -5 \ 0 & 3-\lambda \end{array} \right| \]Expanding the 2x2 determinant:\[ = (-2-\lambda) ((-2-\lambda)(3-\lambda)) = -2-\lambda ((-2-\lambda)(3-\lambda)) \]
04
Simplify the Determinant to Form the Characteristic Polynomial
The expression \((-2-\lambda)((-2-\lambda)(3-\lambda))\) simplifies to get the characteristic polynomial:\[ (-2-\lambda)((-2-\lambda)(3-\lambda)) = (\lambda + 2)(\lambda^2 - \lambda - 6) \]Solving \( \lambda^2 - \lambda - 6 \) by factoring yields:\[(\lambda + 2)((\lambda - 3)(\lambda + 2)) = 0\]So the characteristic polynomial is \((\lambda + 2)^2(\lambda - 3) = 0\).
05
Find Eigenvalues
Setting the characteristic polynomial equal to zero gives the eigenvalues:\[ (\lambda + 2)^2(\lambda - 3) = 0 \]The solutions \( \lambda = -2 \) (with algebraic multiplicity 2) and \( \lambda = 3 \) (with multiplicity 1).
06
Find Eigenvectors for \( \lambda = -2 \)
Substitute \( \lambda = -2 \) back into the equation \( A - \lambda I = 0 \):\( A + 2I = \begin{pmatrix} 0 & 0 & 0 \ -5 & 0 & -5 \ 5 & 0 & 5 \end{pmatrix} \), and solve \( (A + 2I)\mathbf{x} = 0 \).The solution leads to vectors of form \( \begin{pmatrix} x_1 \ 0 \ -x_1 \end{pmatrix} \), a basis for the null space when \( x_1 eq 0 \).
07
Find Eigenvector for \( \lambda = 3 \)
Substitute \( \lambda = 3 \) into the equation \( A - \lambda I = 0 \):\( A - 3I = \begin{pmatrix} -5 & 0 & 0 \ -5 & -5 & -5 \ 5 & 0 & 0 \end{pmatrix} \), and solve \( (A - 3I)\mathbf{x} = 0 \).A solution is \( \begin{pmatrix} 0 \ 1 \ 0 \end{pmatrix} \), a basis vector for the null space.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Polynomial
To find the eigenvalues of a matrix, we must first derive its characteristic polynomial. This is a critical concept in linear algebra that allows us to determine the different perspectives from which our matrix operates in a multi-dimensional space. The characteristic polynomial is formed using the determinant of the difference between the matrix and the scalar multiples of the identity matrix.
Given a matrix \( A \), the process involves subtracting \( \lambda I \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. For instance, with our matrix \( A \), we first construct \( A - \lambda I \), leading to a new matrix form. The characteristic polynomial is then:
Given a matrix \( A \), the process involves subtracting \( \lambda I \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix. For instance, with our matrix \( A \), we first construct \( A - \lambda I \), leading to a new matrix form. The characteristic polynomial is then:
- Derived by calculating the determinant of this new matrix \( A - \lambda I \).
- This determinant results in a polynomial equation in terms of \( \lambda \).
Eigenvectors
Once the eigenvalues are identified through the characteristic polynomial, eigenvectors come into play. An eigenvector is a non-zero vector that only changes by a scalar factor when a linear transformation is applied to it. It aligns with the direction your matrix inherently wants to stretch or compress.
- For an eigenvalue \( \lambda \), an eigenvector \( \mathbf{x} \) satisfies \( A\mathbf{x} = \lambda \mathbf{x} \).
- In simpler terms, the matrix transformation shrinks or stretches \( \mathbf{x} \) along the vector's line.
Matrix Determinant
The matrix determinant is a crucial tool in the determination of the characteristic polynomial. It essentially provides a scalar value that encodes certain properties of the matrix, such as invertibility or the volume scaling factor when the matrix is seen as a linear transformation.
- The determinant is calculated specifically for square matrices.
- The process involves a sum of products of entries from the matrix, observing specific sign rules from the matrix's elements.
Algebraic Multiplicity
Algebraic multiplicity refers to the number of times a particular eigenvalue appears as a root in the characteristic polynomial. It indicates not only the dimension of the eigenvalue's eigenspace but also hints at symmetry and stability within the matrix transformation.
- An eigenvalue’s algebraic multiplicity is determined by the exponent of the factor in the polynomial.
- This multiplicity affects the nature and behavior of the matrix considerably.