Question

Consider a \(2 \times 2\) system \(\mathbf{x}^{\prime}=\mathbf{A} \mathbf{x}\). If we assume that \(r_{1} \neq r_{2}\), the general solution is \(\mathbf{x}=c_{1} \xi^{(1)} e^{t_{1}^{\prime}}+c_{2} \xi^{(2)} e^{\prime 2},\) provided that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent In this problem we establish the linear independence of \(\xi^{(1)}\) and \(\xi^{(2)}\) by assuming that they are linearly dependent, and then showing that this leads to a contradiction. $$ \begin{array}{l}{\text { (a) Note that } \xi \text { (i) satisfies the matrix equation }\left(\mathbf{A}-r_{1} \mathbf{I}\right) \xi^{(1)}=\mathbf{0} ; \text { similarly, note that }} \\ {\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(2)}=\mathbf{0}} \\ {\text { (b) Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right) \xi^{(1)}=\left(r_{1}-r_{2}\right) \mathbf{\xi}^{(1)}} \\\ {\text { (c) Suppose that } \xi^{(1)} \text { and } \xi^{(2)} \text { are linearly dependent. Then } c_{1} \xi^{(1)}+c_{2} \xi^{(2)}=\mathbf{0} \text { and at least }}\end{array} $$ $$ \begin{array}{l}{\text { one of } c_{1} \text { and } c_{2} \text { is not zero; suppose that } c_{1} \neq 0 . \text { Show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=\mathbf{0}} \\ {\text { and also show that }\left(\mathbf{A}-r_{2} \mathbf{I}\right)\left(c_{1} \boldsymbol{\xi}^{(1)}+c_{2} \boldsymbol{\xi}^{(2)}\right)=c_{1}\left(r_{1}-r_{2}\right) \boldsymbol{\xi}^{(1)} \text { . Hence } c_{1}=0, \text { which is }} \\\ {\text { a contradiction. Therefore } \xi^{(1)} \text { and } \boldsymbol{\xi}^{(2)} \text { are linearly independent. }}\end{array} $$ $$ \begin{array}{l}{\text { (d) Modify the argument of part (c) in case } c_{1} \text { is zero but } c_{2} \text { is not. }} \\ {\text { (e) Carry out a similar argument for the case in which the order } n \text { is equal to } 3 \text { ; note that }} \\ {\text { the procedure can be extended to cover an arbitrary value of } n .}\end{array} $$

Short answer

**Question:** Show that the vectors \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent given the matrix equation \((\mathbf{A}-r_{i}\mathbf{I})\xi^{(i)} = \mathbf{0}\) for \(i=1,2\) and \(r_{1} \neq r_{2}\). **Answer:** We assumed that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly dependent and showed that this assumption leads to contradictions in both cases when \(c_{1} \neq 0\) and when \(c_{1} = 0, c_{2} \neq 0\). Therefore, the vectors \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent.

Step-by-step solution

a) Matrix equation for \(\xi^{(1)}\) and \(\xi^{(2)}\)

Given the matrix equation \((\mathbf{A}-r_{i}\mathbf{I})\xi^{(i)} = \mathbf{0}\) for \(i=1,2\).

b) Show that \((\mathbf{A}-r_{2} \mathbf{I}) \xi^{(1)} = (r_{1}-r_{2})\mathbf{\xi}^{(1)}\)

First, multiply both sides of the given matrix equation for \(\xi^{(1)}\) by \(\mathbf{A}-r_{2}\mathbf{I}\): $$(\mathbf{A}-r_{2} \mathbf{I})(\mathbf{A}-r_{1} \mathbf{I})\xi^{(1)} = (\mathbf{A}-r_{2} \mathbf{I})\mathbf{0}$$ Now, we know that \(\xi^{(2)}\) satisfies the equation \((\mathbf{A}-r_{2}\mathbf{I})\xi^{(2)} = \mathbf{0}\), which means $$(\mathbf{A}-r_{1} \mathbf{I})\xi^{(1)} = \xi^{(2)}$$ So, substituting in the above equation, we get $$(\mathbf{A}-r_{2} \mathbf{I})\xi^{(1)} = (r_{1}-r_{2}) \mathbf{\xi}^{(1)}$$

c) Assume \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly dependent

Suppose that \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly dependent. Then, there exist scalars \(c_{1}\) and \(c_{2}\) such that $$c_{1} \xi^{(1)}+c_{2} \xi^{(2)} = \mathbf{0}$$ and at least one of \(c_{1}\) and \(c_{2}\) is nonzero. Assume \(c_{1} \neq 0\). Now, apply the matrix \((\mathbf{A}-r_{2} \mathbf{I})\) to both sides of the equation, $$(\mathbf{A}-r_{2} \mathbf{I})(c_{1} \xi^{(1)}+c_{2} \xi^{(2)}) = (\mathbf{A}-r_{2} \mathbf{I})\mathbf{0} = \mathbf{0}$$ From part (b), we have \((\mathbf{A}-r_{2} \mathbf{I}) \xi^{(1)} = (r_{1}-r_{2}) \mathbf{\xi}^{(1)}\), and since \((\mathbf{A}-r_{2} \mathbf{I})\xi^{(2)} = \mathbf{0}\), we can rewrite the above equation as $$c_{1}(r_{1}-r_{2})\mathbf{\xi}^{(1)} = \mathbf{0}$$ Since \(c_{1} \neq 0\) and \(r_{1} \neq r_{2}\), this implies that \(\xi^{(1)} = \mathbf{0}\), which is a contradiction. Therefore, \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent.

d) Modify the argument for the case \(c_{1} = 0\) and \(c_{2} \neq 0\)

If \(c_{1} = 0\) and \(c_{2} \neq 0\), then the equation \(c_{1} \xi^{(1)}+c_{2} \xi^{(2)} = \mathbf{0}\) becomes $$c_{2} \xi^{(2)} = \mathbf{0}$$ Now, apply the matrix \((\mathbf{A}-r_{1} \mathbf{I})\) to both sides of the equation, $$(\mathbf{A}-r_{1} \mathbf{I})(c_{2} \xi^{(2)}) = (\mathbf{A}-r_{1} \mathbf{I})\mathbf{0} = \mathbf{0}$$ Since \((\mathbf{A}-r_{1} \mathbf{I})\xi^{(1)} = \mathbf{0}\) and \((\mathbf{A}-r_{2} \mathbf{I}) \xi^{(2)} = (r_{1}-r_{2}) \mathbf{\xi}^{(2)}\), this equation becomes $$c_{2}(r_{1}-r_{2})\mathbf{\xi}^{(2)} = \mathbf{0}$$ Since \(c_{2} \neq 0\) and \(r_{1} \neq r_{2}\), this implies that \(\xi^{(2)} = \mathbf{0}\), which is a contradiction. Therefore, \(\xi^{(1)}\) and \(\xi^{(2)}\) are linearly independent.

Key concepts

2x2 Systems of Differential Equations

Differential equations are equations that involve functions and their derivatives. A common class of these equations is the system of linear differential equations, especially noteworthy for applications in various fields. In the context of a 2x2 system, we specifically refer to two linear differential equations organized in a matrix form:
This setup allows us to simultaneously solve these two equations for two unknown functions, typically denoted by a vector \( \mathbf{x}(t) \). For example,

• The differential equation is expressed as \( \mathbf{x}^{\prime} = \mathbf{A} \mathbf{x} \),
• where \( \mathbf{x} \) is a 2-dimensional vector of functions, and \( \mathbf{A} \) is a 2x2 matrix with constant coefficients.

Each component of the vector function is a solution to the differential equation, describing how the system changes over time. Solving such systems involves finding so-called eigenvectors and eigenvalues of the matrix \( \mathbf{A} \), which we'll explore more deeply in a later section.

Matrix Equation

Matrix equations form the backbone of understanding systems of differential equations. They allow us to translate differential equations into a form that can be more easily manipulated and solved.
In linear algebra, we often come across the matrix equation \( \mathbf{A} \mathbf{x} = \mathbf{b} \), but in the context of differential equations, our focus is often on homogeneous equations, specifically the form \( (\mathbf{A} - r_i \mathbf{I}) \xi^{(i)} = \mathbf{0} \). Here,

• \( \mathbf{A} \) is a given matrix, often representing the coefficients of the system.
• \( \xi^{(i)} \) are the eigenvectors.
• \( r_i \) are the eigenvalues, and \( \mathbf{I} \) is the identity matrix.

The purpose of this equation is to identify non-trivial solutions \( \xi^{(i)} \) that satisfy the equation for given \( r_i \). This approach helps us solve systems by finding a particular solution and constructing a general solution.

Contradiction Proof

A contradiction proof is a logical process used to demonstrate that a specific assumption leads to an untenable result, thus proving the original assumption must be false. It’s a handy tool in mathematics when direct proof is complex.
Within the context of proving linear independence of vectors \( \xi^{(1)} \) and \( \xi^{(2)} \) in a system of differential equations, the contradiction arises like this:

• We assume that \( \xi^{(1)} \) and \( \xi^{(2)} \) are linearly dependent, meaning \( c_1 \xi^{(1)} + c_2 \xi^{(2)} = \mathbf{0} \) for some constants \( c_1 \) and \( c_2 \)
• After applying matrix manipulations as per problem requirements, these lead to a conclusion that contradicts our initial assumption.

Such proofs are critical in ensuring that the solutions found are not only mathematically valid but also correspond to realistic and substantive scenarios.

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are pivotal in solving linear systems, including differential equations. These concepts deal with transformations represented by matrices.
When we apply a matrix \( \mathbf{A} \) to a vector \( \xi \), and the resulting vector is a scalar multiple of \( \xi \), this vector is an eigenvector. The corresponding scalar is the eigenvalue.

• Mathematically, this is shown as \( \mathbf{A} \xi = \lambda \xi \), where \( \lambda \) is the eigenvalue.
• To solve \( (\mathbf{A} - \lambda \mathbf{I}) \xi = \mathbf{0} \), we determine eigenvectors \( \xi \) for each \( \lambda \).
• The eigenvalues and eigenvectors help in transforming a complex differential system into a simple form that can be easily solved or analyzed.

In systems of differential equations, these forms aid in constructing the general solution, leading to deeper insights into both the stability and behavior of the system over time.