Question

Rewrite the linear system as a matrix equation \(\mathbf{y}^{\prime}=A \mathbf{y}\), and compute the eigenvalues of the matrix \(A\). $$ \begin{array}{ll} y_{1}^{\prime} & =2 y_{2} \\ y_{2}^{\prime} & =-2 y_{1} \end{array} $$

Short answer

QUESTION: Rewrite the given system of linear ordinary differential equations as a matrix equation and find the eigenvalues of the matrix: y₁' = 2y₂; y₂' = -2y₁; ANSWER: The given system can be rewritten as the matrix equation: y' = Ay, where A = | 0 2 | | -2 0 | The eigenvalues of matrix A are λ = 2i and λ = -2i.

Step-by-step solution

Rewrite the linear system as a matrix equation

To convert the given system into the matrix equation \(\mathbf{y}^{\prime} = A\mathbf{y}\), we consider the vectors \(\mathbf{y}^{\prime} = (y_1^{\prime}, y_2^{\prime})\) and \(\mathbf{y} = (y_1, y_2)\). Then, our goal is to find a matrix \(A\) such that, $$ \mathbf{y}^{\prime} = A\mathbf{y} = \begin{pmatrix} y_1^{\prime} \\ y_2^{\prime} \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}. $$ Looking at the given linear system, we can deduce the matrix \(A\) as follows, $$ A = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}. $$ So our matrix equation becomes, $$ \mathbf{y}^{\prime} = \begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}\mathbf{y}. $$

Compute the eigenvalues of the matrix A

To compute the eigenvalues of the matrix \(A\), we need to solve the characteristic equation \(|A - \lambda I| = 0\) for \(\lambda\), where \(I\) is the identity matrix. First, let's write down \(A - \lambda I\): $$ A - \lambda I = \begin{pmatrix} 0-\lambda & 2 \\ -2 & 0-\lambda \end{pmatrix} =\begin{pmatrix} -\lambda & 2 \\ -2 & -\lambda \end{pmatrix}. $$ Now, let's compute the determinant \(|A - \lambda I|\) and set it to zero, $$ |A - \lambda I| = (-\lambda)(-\lambda) - (2)(-2) = \lambda^2 + 4. $$ Setting \(|A - \lambda I| = 0\) and solving for lambda, we get: $$ \lambda^2 + 4 = 0 \Rightarrow \lambda^2 = -4 \Rightarrow \lambda = \pm 2i. $$ Hence, the eigenvalues of the matrix \(A\) are \(\lambda = 2i\) and \(\lambda = -2i\).

Key concepts

Matrix Equations

Matrix equations are a powerful way to represent linear systems in a compact and clear form. In the given exercise, we started with a linear system of two equations, where

• \(y_1' = 2y_2\)
• \(y_2' = -2y_1\)

To convert this into a matrix equation, we define a vector \(\mathbf{y} = (y_1, y_2)\). Then, we consider the derivative \(\mathbf{y}' = (y_1', y_2')\). Our goal is to find a matrix \(A\) such that the equation \(\mathbf{y}' = A\mathbf{y}\) holds true. By examining the system, we arrive at the matrix \(A\): \[A = \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix}\] Thus, the matrix equation is \(\mathbf{y}' = A\mathbf{y}\). This representation simplifies the analysis of the system and makes computational tasks easier to manage.

Eigenvalues

Eigenvalues are critical in understanding the behavior of linear transformations, such as matrix equations. They tell us how a linear transformation changes vectors that are scaled along certain directions. To find the eigenvalues of a matrix, we solve the characteristic equation.
In our case, for the matrix \[A = \begin{pmatrix} 0 & 2 \ -2 & 0 \end{pmatrix},\] we need to solve \(|A - \lambda I| = 0\), where \(I\) is the identity matrix. The eigenvalues are solutions to this polynomial equation, which leads us to complex numbers: \(\lambda = 2i\) and \(\lambda = -2i\).
These eigenvalues indicate the system has an oscillatory behavior, as they are purely imaginary numbers.

Characteristic Equation

The characteristic equation is an essential concept in linear algebra, providing the foundation for finding eigenvalues of a matrix.
For a given matrix \(A\), the characteristic equation is derived by setting the determinant of \(A - \lambda I\) equal to zero: \[|A - \lambda I| = 0\] In our exercise,\[A - \lambda I = \begin{pmatrix} -\lambda & 2 \ -2 & -\lambda \end{pmatrix}\] and \[|A - \lambda I| = \lambda^2 + 4.\] By solving \(\lambda^2 + 4 = 0\), we find the eigenvalues \(\lambda = \pm 2i\). The characteristic equation not only provides the eigenvalues but also offers insights into the system's dynamic properties; here, it suggests oscillations.

Differential Equations

Differential equations describe how a function changes with respect to its variables, often representing real-world phenomena like motion or growth.
Linear differential equations can be expressed in matrix form to facilitate their analysis. The given exercise provided a system of differential equations, and by converting it to a matrix equation \(\mathbf{y}' = A\mathbf{y}\), it is easier to handle using tools from linear algebra.
This inclusion of eigenvalues helps predict the behavior of solutions over time. Specifically, for matrices with imaginary eigenvalues, the solutions to the differential equations will typically be oscillatory, involving sines and cosines.
Thus, the matrix equation approach simplifies the study and solution of differential equations, offering valuable insights into the system's time evolution.