Question

Find the eigenvalues and eigenvectors of the following matrices. Do some problems by hand to be sure you understand what the process means. Then check your results by computer. $$\left(\begin{array}{rr} 3 & -2 \\ -2 & 0 \end{array}\right)$$

Short answer

Eigenvalues: 4, -1. Eigenvectors: any scalar multiple of \[\begin{pmatrix} -2 \ 1 \end{pmatrix}\] for \[\lambda = 4\], and \[\begin{pmatrix} 1 \ 2 \end{pmatrix}\] for \[\lambda = -1\].

Step-by-step solution

- Set Up the Characteristic Equation

Given matrix \[A = \begin{pmatrix} 3 & -2 \ -2 & 0 \end{pmatrix}\], we need to find the eigenvalues by solving the characteristic equation. The characteristic equation is given by \[\det(A - \lambda I) = 0\], where \[I\] is the identity matrix and \[\lambda\] are the eigenvalues.

- Compute the Determinant

Subtract \[\lambda\] times the identity matrix from \[A\]: \[A - \lambda I = \begin{pmatrix} 3 - \lambda & -2 \ -2 & -\lambda \end{pmatrix}\]. Now compute the determinant: \[\det(A - \lambda I) = (3 - \lambda)(-\lambda) - (-2)(-2)\].

- Simplify the Determinant

Simplify the expression: \[\det(A - \lambda I) = -3\lambda + \lambda^2 - 4 = \lambda^2 - 3\lambda - 4\]. Set this equal to zero to find the eigenvalues: \[\lambda^2 - 3\lambda - 4 = 0\].

- Solve the Characteristic Equation

Solve the quadratic equation \[\lambda^2 - 3\lambda - 4 = 0\] to find the eigenvalues. Factor the equation: \[(\lambda - 4)(\lambda + 1) = 0\]. Therefore, the eigenvalues are \[\lambda_1 = 4\] and \[\lambda_2 = -1\].

- Find Eigenvectors for \(\lambda_1 = 4\)

Substitute \[\lambda_1 = 4\] into \[A - \lambda I\] to get: \[\begin{pmatrix} 3 - 4 & -2 \ -2 & 0 - 4 \end{pmatrix} = \begin{pmatrix} -1 & -2 \ -2 & -4 \end{pmatrix}\]. Solve \[\begin{pmatrix} -1 & -2 \ -2 & -4 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\] to find the eigenvector.

- Solve for Eigenvector Components

From the first row, \[-x - 2y = 0\], which simplifies to \[x = -2y\]. The eigenvector corresponding to \[\lambda_1 = 4\] is any scalar multiple of \[\begin{pmatrix} -2 \ 1 \end{pmatrix}\].

- Find Eigenvectors for \(\lambda_2 = -1\)

Substitute \[\lambda_2 = -1\] into \[A - \lambda I\] to get: \[\begin{pmatrix} 3 + 1 & -2 \ -2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & -2 \ -2 & 1 \end{pmatrix}\]. Solve \[\begin{pmatrix} 4 & -2 \ -2 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}\] to find the eigenvector.

- Solve for Eigenvector Components

From the first row, \[4x - 2y = 0\] which simplifies to \[2x = y\]. The eigenvector corresponding to \[\lambda_2 = -1\] is any scalar multiple of \[\begin{pmatrix} 1 \ 2 \end{pmatrix}\].

Key concepts

characteristic equation

The characteristic equation is essential for finding the eigenvalues of a matrix. Given a square matrix \[A\], the characteristic equation is formed by taking the determinant of the matrix \[(A - \lambda I)\]. Here, \[\lambda\] represents the eigenvalues, and \[I\] is the identity matrix of the same dimension as \[A\]. Solving the characteristic equation \[(\det(A - \lambda I) = 0)\], we identify the eigenvalues \[\lambda\]. This foundational step transforms the matrix into a polynomial equation, which we solve to find eigenvalues.

determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It provides essential insights about the matrix itself, such as whether it is invertible or singular. In the context of finding eigenvalues, calculating the determinant of \[(A - \lambda I)\] reduces the matrix into a solvable polynomial form. For example:

\[\det \begin{pmatrix} 3 - \lambda & -2 \ -2 & -\lambda \end{pmatrix} = (3 - \lambda)(-\lambda) - (-2)(-2)\].

By simplifying this, we contribute to forming the characteristic polynomial equation.

quadratic equation

A quadratic equation is a second-order polynomial equation, usually in the format \[ax^2 + bx + c = 0\]. In our exercise, once we compute the determinant, we get a characteristic polynomial in this form. For example:

\[\lambda^2 - 3\lambda - 4 = 0\].

To solve for \[\lambda\] (the eigenvalues), we can factorize this quadratic equation:

\[(\lambda - 4)(\lambda + 1) = 0\].

The solutions \[\lambda = 4\] and \[\lambda = -1\] are the eigenvalues of our matrix.

linear algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear transformations. The process of finding eigenvalues and eigenvectors is a crucial application of linear algebra.

Eigenvalues represent the scalar values that, when multiplied by their corresponding eigenvectors, transform the eigenvectors in a specific manner without altering their direction. In the matrix equation \[A\mathbf{x} = \lambda\mathbf{x}\], \[\lambda\] is an eigenvalue, and \[\mathbf{x}\] is the corresponding eigenvector.

To find these, we often use steps from linear algebra like forming and solving characteristic equations and computing determinants. This journey provides deep insights into the properties and behaviors of linear transformations.