Question

Find a fundamental system of solations of Eq. (1), where $$ A=\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) $$

Short answer

Answer: The fundamental system of solutions for the given equation is: $$ \left\lbrace \left(\begin{matrix} a \\ ai \end{matrix}\right), \left(\begin{matrix} b \\ -bi \end{matrix}\right) \right\rbrace $$ Where a and b are real numbers.

Step-by-step solution

Find the characteristic equation

To find the characteristic equation of the given matrix A, we need to calculate the determinant of (A - λI), where λ is the eigenvalue and I is the identity matrix. In our case, A is: $$ A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} $$ and λI is: $$ \lambda I = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} $$ Now, let's find (A - λI): $$ A - \lambda I = \begin{pmatrix} 0-\lambda & 1 \\ -1 & 0-\lambda \end{pmatrix} = \begin{pmatrix} -\lambda & 1 \\ -1 & -\lambda \end{pmatrix} $$ The characteristic equation is the determinant of (A - λI) equals zero, which we calculate below: $$ \text{det}(A - \lambda I) = (-\lambda)(-\lambda) - (1)(-1) = \lambda^2 + 1 $$ We set the determinant equal to zero: $$ \lambda^2 + 1 = 0 $$

Find the eigenvalues

To find the eigenvalues, we need to solve the characteristic equation we found in step 1: $$ \lambda^2 + 1 = 0 \Rightarrow \lambda^2 = -1 $$ To get eigenvalues, we solve for λ: $$ \lambda = \pm\sqrt{-1} = \pm i $$ So the eigenvalues are λ₁ = i and λ₂ = -i.

Find eigenvectors

Now that we have the eigenvalues, we need to find the corresponding eigenvectors for each eigenvalue by solving the following equation: $$ (A - \lambda I) \vec{v} = \vec{0} $$ For λ₁ = i: $$ \begin{pmatrix} -i & 1 \\ -1 & -i\end{pmatrix} \vec{v} = \vec{0} $$ We can rewrite this system of equations as: $$ (-i)v_1 + v_2 = 0 $$ $$ -v_1 - iv_2= 0 $$ Let v₁ = a -> v₂ = ai (a ∈ ℝ). So the eigenvector for λ₁ = i is: $$ \vec{v}_1 = \begin{pmatrix} a \\ ai \end{pmatrix} $$ For λ₂ = -i: $$ \begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix} \vec{v} = \vec{0} $$ We can rewrite this system of equations as: $$ (iv_1) + v_2 = 0 $$ $$ (-v_1) + (iv_2)= 0 $$ Let v₁ = b -> v₂ = -bi (b ∈ ℝ). So the eigenvector for λ₂ = -i is: $$ \vec{v}_2 = \begin{pmatrix} b \\ -bi \end{pmatrix} $$

Express the fundamental system of solutions

Finally, we express the fundamental system of solutions using the eigenvectors we found. A fundamental system of solutions for the given equation is: $$ \left\lbrace \left(\begin{matrix} a \\ ai \end{matrix}\right), \left(\begin{matrix} b \\ -bi \end{matrix}\right) \right\rbrace $$ Where a and b are real numbers.

Key concepts

Characteristic Equation Explained

Understanding the characteristic equation is essential when studying the behavior of matrices, particularly in the context of eigenvalues and eigenvectors. It is essentially a polynomial equation derived from the matrix in question, specifically formed by the determinant of the matrix subtracted by an eigenvalue times the identity matrix. Following the format \[\text{det}(A - \lambda I) = 0\], we find a characteristic equation that will help us determine the eigenvalues.In our exercise, the matrix A has a simple structure with zeroes on the diagonal and ones off-diagonal. This characteristic equation exemplifies how such a matrix's properties translate into a quadratic equation, \(\lambda^2 + 1 = 0\), which remarkably leads to complex solutions. The presence of complex solutions is not unusual and indicates the kinds of dynamics that the matrix A can produce in systems like linear differential equations.

Digging into Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental to understanding linear transformations represented by matrices. An eigenvalue, \(\lambda\), is a special scalar that, when multiplied by a corresponding non-zero eigenvector, \(\vec{v}\), yields the same result as if you transformed \(\vec{v}\) by the matrix. More formally, for a matrix A, this relationship is represented by \[A\vec{v} = \lambda\vec{v}\].

The process of finding eigenvalues was demonstrated in the exercise, leading us to complex numbers, \(\lambda = \pm i\), due to the nature of our specific matrix A. These eigenvalues further inform us about the system's behavior, such as oscillatory motion commonly seen in systems modeled by matrices with complex eigenvalues.Once the eigenvalues are found, we then proceed to find the eigenvectors, which form the basis for the space transformed by the matrix. In our case, each eigenvalue provided a vector that is scaled and rotated, illustrating the unique way matrix A interacts with vectors in its space.

Determinant of a Matrix

The determinant of a matrix is a scalar value that provides important information about the matrix. It can tell us if the matrix is invertible, its scaling factor in linear transformations, and contributes to finding eigenvalues. In broader terms, the determinant of a square matrix is calculated from the sum and products of its elements according to specific rules.

The exercise showcases the calculation of the determinant by subtracting \(\lambda\) times the identity matrix from A. The resulting polynomial equation was set to zero, which is the condition that defines eigenvalues. It's interesting to note that the determinant is the foundation for defining the characteristic equation, highlighting its significance in linear algebra. For our 2x2 matrix A, the determinant unveils the complex nature of the solutions in the system it represents.

Navigating Complex Eigenvalues

Complex eigenvalues usually emerge in systems with rotational and oscillatory motions and are common in the study of differential equations. Given a real square matrix, such as in our example problem, the presence of complex eigenvalues suggests that the matrix can describe such motions. This is particularly true in systems of linear differential equations where the solutions describe dynamic processes.

In the provided exercise, the complex eigenvalues \(\lambda = \pm i\) arise from the quadratic characteristic equation. These complex solutions lead to complex eigenvectors, which correspond to sine and cosine waves when applied in the context of systems of differential equations. These waves reflect the oscillatory nature of the system that the matrix A can represent.

System of Linear Differential Equations

A system of linear differential equations consists of multiple equations that involve the derivatives of several unknown functions related to each other through linear expressions. These types of systems frequently appear in modeling real-world processes, where the interaction between different quantities can be approximated by linear relationships.

Our matrix A represents a system of two linear differential equations with constant coefficients, and the solutions are sought in the form of eigenvectors. The fundamental system of solutions built from these eigenvectors provides a concise way to describe the general solution to the system. In our case, the complex eigenvalues and associated eigenvectors imply that the general solution will involve trigonometric functions reflective of the system's inherent oscillatory nature. This set of solutions is what we look for when solving such differential equations and helps us to understand and predict the system's behavior over time.