Problem 22 - Differential Equations and Linear Algebra

Question

Let \(\lambda=a+i b, b \neq 0,\) be an eigenvalue of the \(n \times n\) (real) matrix \(A\) with corresponding eigenvector \(\mathbf{v}=\mathbf{r}+i \mathbf{s} .\) Then we have shown in the text that two real-valued solutions to \(\mathbf{x}^{\prime}=A \mathbf{x}\) are $$ \begin{array}{l} \mathbf{x}_{1}(t)=e^{a t}[\cos b t \mathbf{r}-\sin b t \mathbf{s}] \\ \mathbf{x}_{2}(t)=e^{a t}[\sin b t \mathbf{r}+\cos b t \mathbf{s}] \end{array} $$ Prove that \(\mathbf{x}_{1}\) and \(\mathbf{x}_{2}\) are linearly independent on any interval. (You may assume that \(\mathbf{r}\) and \(\mathbf{s}\) are linearly independent in \(\mathbb{R}^{n} .\) ) The remaining problems in this section investigate general properties of solutions to \(\mathbf{x}^{\prime}=A \mathbf{x},\) where \(A\) is a nondefective matrix.

Step-by-Step Solution

Answer

To prove that the two real-valued solutions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) are linearly independent on any interval, we assumed a linear combination of the solutions and showed that the coefficients must be zero. Specifically, we found that if \(C_1\mathbf{x}_1(t) + C_2\mathbf{x}_2(t) = 0\), both constants \(C_1\) and \(C_2\) must be zero for all \(t\). This proves their linear independence.

Step 1: Assume a linear combination with constant coefficients of \(\mathbf{x}_1\) and \(\mathbf{x}_2\)

Let's assume that \(C_1\mathbf{x}_1(t) + C_2\mathbf{x}_2(t) = 0\) for some constants \(C_1\) and \(C_2\). We have: \(C_1e^{at}[\cos(bt)\mathbf{r}-\sin(bt)\mathbf{s}] + C_2e^{at}[\sin(bt)\mathbf{r}+\cos(bt)\mathbf{s}] = 0\) Now, let's distribute the constants and combine the terms with \(\mathbf{r}\) and \(\mathbf{s}\) separately.

Step 2: Combine terms with \(\mathbf{r}\) and \(\mathbf{s}\)

After distributing and combining the terms, we get: \([C_1\cos(bt) + C_2\sin(bt)]e^{at}\mathbf{r} + [-C_1\sin(bt) + C_2\cos(bt)]e^{at}\mathbf{s} = 0\) In order for the above equation to hold true for all \(t\), both coefficients of \(\mathbf{r}\) and \(\mathbf{s}\) must be zero.

Step 3: Set coefficients of \(\mathbf{r}\) and \(\mathbf{s}\) to zero

We have two equations: 1. \(C_1\cos(bt) + C_2\sin(bt) = 0\) 2. \(-C_1\sin(bt) + C_2\cos(bt) = 0\)

Step 4: Show that constants \(C_1\) and \(C_2\) must be zero

Since it needs to be zero for all \(t\), let's consider two cases: 1. When \(t=0\), \(C_1 = 0\) and \(-C_2 = 0\). 2. When \(t=\frac{\pi}{2b}\), (it's allowed since \(b\neq0\)) \(C_1 = 0\) and \(C_2 = 0\). In both cases, we get that both constants, \(C_1\) and \(C_2\), must be zero. Thus, we have proved that the two real-valued solutions \(\mathbf{x}_1\) and \(\mathbf{x}_2\) are linearly independent on any interval.

Key Concepts

Eigenvalues and Eigenvectors

Understanding the concept of eigenvalues and eigenvectors is fundamental in various areas of mathematics and science, including the study of linear differential equations. An eigenvalue of a matrix is a scalar \(\lambda\) which, when used to multiply an eigenvector, does not change the direction of that vector, only its magnitude or potentially its direction if the scalar is negative. Formally, if \(A\) is a square matrix, \(\mathbf{v}\) is a non-zero vector and \(\lambda\) is a scalar such that \(A \mathbf{v} = \lambda \mathbf{v}\), then \(\lambda\) is the eigenvalue and \(\mathbf{v}\) is the corresponding eigenvector.

Eigenvectors are particularly important because they can vastly simplify the process of analyzing systems, for example, being the basis for solving systems of linear differential equations. When considering the eigenvalues of a matrix, they can be real or complex. Real eigenvalues lead to solutions that are straightforward, while complex eigenvalues introduce oscillatory behavior due to their trigonometric components.

Linear Differential Equations

A linear differential equation is an equation involving an unknown function and its derivatives. In the system \(\mathbf{x}'=A\mathbf{x}\), we are dealing with a first-order linear differential equation, where \(A\) is an \(n \times n\) constant matrix and \(\mathbf{x}\) is an unknown vector of functions. Such equations often describe dynamic systems, for instance, in physics and engineering.

One powerful method of solving these equations is through the use of eigenvalues and eigenvectors, which can transform a system of equations into a set of scalar differential equations. This occurs because the solution to the system \(\mathbf{x}'=A\mathbf{x}\) involves expressions of the form \(e^{\lambda t}\), where \(\lambda\) is the eigenvalue. This approach reduces the problem of solving a matrix system to solving simpler, independent scalar equations for each eigenvalue and its corresponding eigenvector.

Complex Eigenvalues

Complex eigenvalues arise naturally when working with some matrices, particularly in systems that exhibit oscillatory behavior. These are values of \(\lambda\) in the form \(a + ib\), where \(i\) is the imaginary unit and \(b eq 0\). The complex nature of the eigenvalue leads to exponents in the solutions that are trigonometric functions, as seen in the exercise.

When a matrix \(A\) has complex eigenvalues, solutions to the system of linear differential equations typically involve a combination of exponential growth or decay (governed by the real part \(a\)) and oscillatory motion (governed by the imaginary part \(ib\)). Complex eigenvalues always come in conjugate pairs, and so do their eigenvectors, meaning if you have an eigenvalue \(a + ib\), there will also be an eigenvalue \(a - ib\), and the associated eigenvectors will likewise be complex conjugates of each other.

Real-Valued Solutions

Although complex eigenvalues imply complex-valued solutions, it's important to note that we can still extract real-valued solutions from these complex expressions. This is particularly relevant for physical systems where real-valued functions are often required to describe real-world phenomena.

In the exercise, two linearly independent real-valued solutions \(\mathbf{x}_1(t)\) and \(\mathbf{x}_2(t)\) were demonstrated using combinations of trigonometric functions and the real and imaginary parts of an eigenvector associated with a complex eigenvalue. Because the original system is real, the complex parts of the solution can be combined in such a way that they cancel out, leaving only real functions. This allows us to interpret the solutions in a real context, even if the underlying eigenvalues and eigenvectors are complex.