Problem 11 - Differential Equations and Linear Algebra

Question

The matrix \(A=\left[\begin{array}{rrrr}0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0\end{array}\right]\) has characteristic polynomial \(p(\lambda)=\left(\lambda^{2}+1\right)^{2} .\) Determine two complex-valued solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\) of the form \(\mathbf{x}=e^{A t} \mathbf{v},\) and hence, find four linearly independent real-valued solutions to the differential system.

Step-by-Step Solution

Answer

The eigenvalues of the given matrix A are λ = i and λ = -i. The corresponding eigenvectors are \(\mathbf{v}_1=\left( \begin{array}{c} i \\ 1 \\ 0 \\ 1 \end{array} \right)\) and \(\mathbf{v}_2=\left( \begin{array}{c} -i \\ 1 \\ 0 \\ 1 \end{array} \right)\), respectively. Hence, the complex-valued solutions to the differential equation system are \(\mathbf{x}_1(t)=\left( \begin{array}{c} ie^{it} \\ e^{it} \\ 0 \\ e^{it} \end{array} \right)\) and \(\mathbf{x}_2(t)=\left( \begin{array}{c} -ie^{-it} \\ e^{-it} \\ 0 \\ e^{-it} \end{array} \right)\). From these complex-valued solutions, we can find four linearly independent real-valued solutions: \(\mathbf{x}_3(t)=\left( \begin{array}{c} \cos(t) \\ \sin(t) \\ 0 \\ \sin(t) \end{array} \right)\), \(\mathbf{x}_4(t)=\left( \begin{array}{c} \sin(t) \\ -\cos(t) \\ 0 \\ -\cos(t) \end{array} \right)\), \(\mathbf{x}_5(t)=\left( \begin{array}{c} \cos(t) \\ \sin(t) \\ 0 \\ \sin(t) \end{array} \right)\), and \(\mathbf{x}_6(t)=\left( \begin{array}{c} -\sin(t) \\ \cos(t) \\ 0 \\ \cos(t) \end{array} \right)\).

Step 1: Find the eigenvalues

Given the characteristic polynomial \(p(\lambda)=(\lambda^2+1)^2\), we can find the eigenvalues by setting the polynomial to zero and solving for λ: \[(\lambda^2+1)^2=0\] The solutions to this equation are λ = i and λ = -i, where i is the imaginary unit.

Step 2: Find eigenvectors for each eigenvalue

For each eigenvalue, find an eigenvector by solving \((A-\lambda I)\mathbf{v}=0\). For λ = i: \((A-iI)\mathbf{v} = 0\) \[\left( \begin{array}{cccc} -i & -1 & 0 & 0 \\ 1 & -i & 0 & 0 \\ 1 & 0 & -i & -1 \\ 0 & 1 & 1 & -i \end{array} \right)\mathbf{v}=0\] The augmented matrix of the system can be row reduced to: \[\left( \begin{array}{cccc} 1 & -i & 0 & 0 \\ 0 & 1 & 0 & -i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)\] Thus, the eigenvector corresponding to λ = i is: \[\mathbf{v}_1=\left( \begin{array}{c} i \\ 1 \\ 0 \\ 1 \end{array} \right)\] For λ = -i: \((A+iI)\mathbf{v} = 0\) \[\left( \begin{array}{cccc} i & -1 & 0 & 0 \\ 1 & i & 0 & 0 \\ 1 & 0 & i & -1 \\ 0 & 1 & 1 & i \end{array} \right)\mathbf{v}=0\] The augmented matrix of the system can be row reduced to: \[\left( \begin{array}{cccc} 1 & i & 0 & 0 \\ 0 & 1 & 0 & i \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right)\] Thus, the eigenvector corresponding to λ = -i is: \[\mathbf{v}_2=\left( \begin{array}{c} -i \\ 1 \\ 0 \\ 1 \end{array} \right)\]

Step 3: Find complex-valued solutions

Given that we have found eigenvectors corresponding to the eigenvalues, we can now form the complex-valued solutions to the differential equation system: \[\mathbf{x}_1(t)=e^{i t}\mathbf{v}_1=\left( \begin{array}{c} ie^{it} \\ e^{it} \\ 0 \\ e^{it} \end{array} \right)\] \[\mathbf{x}_2(t)=e^{-i t}\mathbf{v}_2=\left( \begin{array}{c} -ie^{-it} \\ e^{-it} \\ 0 \\ e^{-it} \end{array} \right)\]

Step 4: Find real-valued solutions

From the complex-valued solutions above, we can find four linearly independent real-valued solutions: \[\mathbf{x}_3(t)=\Re(\mathbf{x}_1(t))=\left( \begin{array}{c} \cos(t) \\ \sin(t) \\ 0 \\ \sin(t) \end{array} \right)\] \[\mathbf{x}_4(t)=\Im(\mathbf{x}_1(t))=\left( \begin{array}{c} \sin(t) \\ -\cos(t) \\ 0 \\ -\cos(t) \end{array} \right)\] \[\mathbf{x}_5(t)=\Re(\mathbf{x}_2(t))=\left( \begin{array}{c} \cos(t) \\ \sin(t) \\ 0 \\ \sin(t) \end{array} \right)\] \[\mathbf{x}_6(t)=\Im(\mathbf{x}_2(t))=\left( \begin{array}{c} -\sin(t) \\ \cos(t) \\ 0 \\ \cos(t) \end{array} \right)\] Since the real parts of both complex-valued solutions are the same, we have four linearly independent real-valued solutions: \(\mathbf{x}_3(t), \mathbf{x}_4(t), \mathbf{x}_5(t)\) and \(\mathbf{x}_6(t)\).

Key Concepts

Characteristic Polynomial

The characteristic polynomial is a cornerstone concept in linear algebra that plays a pivotal role in understanding the behavior of linear transformations and matrices. It is particularly crucial when dealing with differential equations, as it helps determine the eigenvalues of a matrix. The characteristic polynomial of a square matrix A is defined by determinant(A - λI), where I is the identity matrix and λ represents the eigenvalues we are solving for.

For instance, the given matrix in the exercise results in the polynomial (λ2 + 1)2. Setting this polynomial equal to zero and solving it provides us with the eigenvalues, which are the roots of the characteristic polynomial. In this particular case, it results in complex eigenvalues λ = i and λ = -i. These eigenvalues are then used to determine the behavior of solutions to the corresponding differential equation.

Eigenvalues and Eigenvectors

Understanding eigenvalues and eigenvectors is fundamental in the study of linear systems and differential equations. An eigenvalue is a scalar that indicates how much a corresponding eigenvector is stretched or squished during a linear transformation. An eigenvector is a non-zero vector that only changes by a scalar factor when that linear transformation is applied.

Once the eigenvalues are found by solving the characteristic polynomial, as in the given example, we proceed to find the associated eigenvector by solving the equation (A - λI)v = 0. In the context of differential equations, every eigenvalue-eigenvector pair helps construct a solution to the differential system. These solutions are of vital importance as they describe the evolution of the system over time.

Complex-valued Solutions

When dealing with differential equations, we sometimes encounter complex-valued solutions. These are solutions that include complex numbers, typically arising from complex eigenvalues. They have a form x(t) = eAtv, where e^At represents the matrix exponential of A multiplied by time t, and v is the corresponding eigenvector.

In the exercise, the solutions are expressed as x1(t) and x2(t), which are complex due to the complex eigenvalues i and -i. These solutions are crucial, as they help us understand the behavior of the system in the complex plane, and they set the stage for finding real-world, real-valued solutions.

Real-valued Solutions

While complex-valued solutions provide deep insights into the behavior of differential systems, in many real-world applications, we are interested in real-valued solutions. These are solutions that do not involve complex numbers and can be derived from complex-valued solutions. Given a pair of complex conjugate eigenvalues and their corresponding eigenvectors, we can use Euler's formula to extract real and imaginary parts, which represent the real-valued solutions.

In the given exercise, the real parts and the imaginary parts of the complex solutions x1(t) and x2(t) yield x3(t), x4(t), x5(t), and x6(t). These functions are sinusoidal in nature and reflect the oscillatory behavior of many physical systems. Since we are interested in linearly independent solutions, we select a set of four that span the solution space, allowing us to cover all possible real-world behaviors of the differential system.