Question

Find the eigenvalues and eigenvectors of the matrices in the following problems. $$\left(\begin{array}{rr}1 & 0 \\\3 & -2\end{array}\right)$$

Short answer

Eigenvalues: 1, -2; Eigenvectors: \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \) for 1, \( \begin{pmatrix} 0 \ 1 \end{pmatrix} \) for -2

Step-by-step solution

- Write the matrix

Given matrix:\[ A = \begin{pmatrix} 1 & 0 \ 3 & -2 \end{pmatrix} \]

- Set up the characteristic equation

To find the eigenvalues, solve the characteristic equation \[ \det(A - \lambda I) = 0 \] where \(I\) is the identity matrix and \(\lambda\) is an eigenvalue. \[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 0 \ 3 & -2 - \lambda \end{pmatrix} \]

- Calculate the determinant

Calculate the determinant of \( A - \lambda I \):\[ \det(A - \lambda I) = (1 - \lambda)(-2 - \lambda) - 3(0) \]Simplify the expression: \[ \det(A - \lambda I) = \lambda^2 + \lambda - 2 \]

- Solve the characteristic equation

Solve the quadratic equation \( \lambda^2 + \lambda - 2 = 0 \):\( \lambda^2 + \lambda - 2 = (\lambda - 1)(\lambda + 2) = 0 \)Thus, the eigenvalues are: \( \lambda_1 = 1 \) and \( \lambda_2 = -2 \)

- Find eigenvector for \( \lambda = 1 \)

Substitute \( \lambda = 1 \) into \( (A - \lambda I) \mathbf{x} = 0 \):\[ \begin{pmatrix} 1 - 1 & 0 \ 3 & -2 - 1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]Which simplifies to:\[ \begin{pmatrix} 0 & 0 \ 3 & -3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]This results in the equations:\[ 0x_1 + 0x_2 = 0 \ 3x_1 - 3x_2 = 0 \]Thus, \( x_1 = x_2 \). The eigenvector corresponding to \( \lambda = 1 \) can be written as any non-zero scalar multiple of \( \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

- Find eigenvector for \( \lambda = -2 \)

Substitute \( \lambda = -2 \) into \( (A - \lambda I) \mathbf{x} = 0 \):\[ \begin{pmatrix} 1 + 2 & 0 \ 3 & -2 + 2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]Which simplifies to:\[ \begin{pmatrix} 3 & 0 \ 3 & 0 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \]This results in the equations:\[ 3x_1 = 0 \ 3x_1 = 0 \]Thus, \( x_1 = 0 \). The eigenvector corresponding to \( \lambda = -2 \) is any non-zero scalar multiple of \( \begin{pmatrix} 0 \ 1 \end{pmatrix} \).

Key concepts

Characteristic Equation

Understanding the characteristic equation is crucial for finding eigenvalues and eigenvectors. In the context of linear algebra, for a square matrix \(A\), the characteristic equation is derived from the determinant of the matrix \((A - \lambda I)\), where \(\lambda\) represents the eigenvalues and \(I\) is the identity matrix.

For our matrix \(A = \begin{pmatrix} 1 & 0 \ 3 & -2 \end{pmatrix}\), we set up the equation as \(\det(A - \lambda I) = 0\). This translates to finding the determinant of the matrix \(\begin{pmatrix} 1 - \lambda & 0 \ 3 & -2 - \lambda \end{pmatrix}\).

Solving this determinant gives us a polynomial, typically quadratic for 2x2 matrices, known as the characteristic equation. In this specific case, solving the equation \(\det \begin{pmatrix} 1 - \lambda & 0 \ 3 & -2 - \lambda \end{pmatrix} = \lambda^2 + \lambda - 2 = 0\) provides us with the eigenvalues.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and linear transformations. It is foundational for understanding the behavior of linear systems.

One of the core objectives in linear algebra is to find eigenvalues and eigenvectors of matrices. Eigenvalues are scalars that indicate the factor by which the eigenvector is scaled during the transformation represented by the matrix.

Given our matrix \(A\), the eigenvalues we find using the characteristic equation represent the scalars by which the original space is stretched or compressed. Solving the characteristic equation is a fundamental step in various applications, including stability analysis, quantum mechanics, and dynamics.

Determinants

Determinants provide a scalar value that summarizes certain properties of a square matrix. They are useful in solving linear equations, understanding matrix properties, and are central to the characteristic equation.

For a given matrix \(A = \begin{pmatrix} 1 & 0 \ 3 & -2 \end{pmatrix}\), the determinant is found by calculating \(\det(A - \lambda I) = (1 - \lambda)(-2 - \lambda) - 3(0)\), resulting in \(\lambda^2 + \lambda - 2\).

A determinant of zero implies that the matrix is singular and does not have an inverse. For our characteristic equation, factoring the quadratic gives us \(\lambda^2 + \lambda - 2 = (\lambda - 1)(\lambda + 2) = 0\). The solutions to this equation, \(\lambda = 1\) and \(\lambda = -2\), are the eigenvalues.

Matrix Transformations

Matrix transformations refer to how matrices alter the space they act upon. In the case of eigenvalues and eigenvectors, we observe how a matrix transforms these vectors under multiplication.

For example, consider \(A = \begin{pmatrix} 1 & 0 \ 3 & -2 \end{pmatrix}\). When we multiply \(A\) by one of its eigenvectors, say \(\mathbf{x} = \begin{pmatrix} 1 \ 1 \end{pmatrix}\) for \(\lambda = 1\), the result will be a scaled version of that eigenvector.

Specifically, \(A\mathbf{x} = \mathbf{x}\) for \(\lambda = 1\), and \(A\mathbf{x} = -2\mathbf{x}\) for \(\lambda = -2\). This relationship succinctly summarizes how the matrix transformation preserves the direction of the eigenvectors while scaling them by their corresponding eigenvalues.