Question

For the given linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\), (a) Compute the eigenpairs of the coefficient matrix \(A\). (b) For each eigenpair found in part (a), form a solution of \(\mathbf{y}^{\prime}=A \mathbf{y}\). (c) Does the set of solutions found in part (b) form a fundamental set of solutions? $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 3 & 2 \\ -9 & -6 \end{array}\right] \mathbf{y} $$

Short answer

Question: Determine if the set of solutions for the given linear system of differential equations is a fundamental set of solutions. Matrix A: $$\left[\begin{array}{rr} 3 & 2 \\ -9 & -6 \end{array}\right]$$ Solution: The set of solutions to the given linear system of differential equations is a fundamental set of solutions.

Step-by-step solution

Compute the eigenpairs of the coefficient matrix A

To find the eigenvalues, we first find the characteristic equation of the matrix A: $$\det(A - \lambda I) = 0$$ where \(I\) is the identity matrix. We can compute the eigenvalues by solving this equation for \(\lambda\): $$\begin{vmatrix} 3-\lambda & 2 \\ -9 & -6-\lambda \end{vmatrix} =0$$ Expanding the determinant, we get: $$(3-\lambda)(-6-\lambda)-(-9 \cdot 2)=\lambda^{2}+\lambda$ Ren=-9$$ We now find the roots of the polynomial: $$\lambda^{2}+\lambda=-\lambda(\lambda+1)=0$$ The eigenvalues are \(\lambda=0\) and \(\lambda=-1\). Next, we find the eigenvectors for both eigenvalues. For \(\lambda = 0\), solve the following equation: $$(A - 0I)\mathbf{v}=0$$ Substitute the given matrix A and \(\lambda = 0\): $$\left[\begin{array}{rr} 3 & 2 \\ -9 & -6 \end{array}\right]\mathbf{v} =\mathbf{0}$$ The eigenvector corresponding to \(\lambda = 0\) is \(\mathbf{v} = \begin{bmatrix}2 \\ -3\end{bmatrix}\). For \(\lambda = -1\), solve the following equation: $$(A - (-1)I)\mathbf{v}=0$$ Substitute the given matrix A and \(\lambda = -1\): $$\left[\begin{array}{rr} 4 & 2 \\ -9 & -5 \end{array}\right]\mathbf{v} =\mathbf{0}$$ The eigenvector corresponding to \(\lambda = -1\) is \(\mathbf{v} = \begin{bmatrix}1 \\ -\frac{9}{2}\end{bmatrix}\). Thus, the eigenpairs are: \((0, \begin{bmatrix}2 \\ -3\end{bmatrix})\) and \((-1, \begin{bmatrix}1 \\ -\frac{9}{2}\end{bmatrix})\).

Form a solution for each eigenpair

Now that we have the eigenpairs, we can write the general solution to the system of differential equations as: $$\mathbf{y}(t) = C_1 e^{\lambda_1 t} \mathbf{v}_1 + C_2 e^{\lambda_2 t} \mathbf{v}_2$$ Plug in the eigenvalues and eigenvectors obtained in the previous step: $$\mathbf{y}(t) = C_1 e^{0t}\begin{bmatrix}2 \\ -3\end{bmatrix} + C_2 e^{-1t}\begin{bmatrix}1 \\ -\frac{9}{2}\end{bmatrix}$$ Simplify the expression: $$\mathbf{y}(t) = C_1 \begin{bmatrix}2 \\ -3\end{bmatrix} + C_2 e^{-t}\begin{bmatrix}1 \\ -\frac{9}{2}\end{bmatrix}$$ The above equation represents the solution for each eigenpair.

Check if the set of solutions form a fundamental set

For the set of solutions to be a fundamental set, they must be linearly independent. Let's verify this using the Wronskian. The Wronskian of \(n\) solutions is given by: $$W(t) = \det\begin{bmatrix}\mathbf{y}_1(t) & \mathbf{y}_2(t)\end{bmatrix}$$ Here, \(\mathbf{y}_1(t) = \begin{bmatrix}2 \\ -3\end{bmatrix}\) and \(\mathbf{y}_2(t) = e^{-t}\begin{bmatrix}1 \\ -\frac{9}{2}\end{bmatrix}\). Calculate the Wronskian: $$W(t) = \det\begin{bmatrix} 2 & e^{-t} \\ -3 & -\frac{9}{2}e^{-t} \end{bmatrix} = -6e^{-t} + 3e^{-t}$$ Since the Wronskian is nonzero for all \(t\), the set of solutions form a fundamental set.

Key concepts

Eigenvalues and Eigenvectors

In the study of linear differential systems, understanding eigenvalues and eigenvectors is crucial. They provide insight into the behavior of systems represented by matrices. For a square matrix \(A\), eigenvalues are numbers \(\lambda\) such that there exists a non-zero vector \(\mathbf{v}\) (the eigenvector), satisfying the equation \(A\mathbf{v} = \lambda\mathbf{v}\). These values indicate scaling factors along specific directions in the matrix's vector space.

To calculate eigenvalues, we first determine the characteristic equation. Eigenvectors are then found by solving the equation \((A - \lambda I)\mathbf{v} = 0\), where \(I\) is the identity matrix. Each eigenvalue has a corresponding eigenvector or vectors. For this particular exercise, we discovered two eigenpairs: \((0, \begin{bmatrix}2 \ -3\end{bmatrix})\) and \((-1, \begin{bmatrix}1 \ -\frac{9}{2}\end{bmatrix})\). These tell us how solutions grow or shrink over time, essential for understanding dynamic systems.

Characteristic Equation

The characteristic equation is at the heart of finding eigenvalues for a matrix. For a matrix \(A\), the characteristic equation is derived from \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix of the same size as \(A\). The determinant \(\det\) results in a polynomial equation where the variable is \(\lambda\).

Solving this polynomial gives all potential eigenvalues for the matrix. In our exercise, the characteristic equation was determined as \(\lambda^2 + \lambda = 0\). Solving it, we identified the eigenvalues \(\lambda = 0\) and \(\lambda = -1\). Each root corresponds to a specific behavior pattern for system solutions, helping predict stability and response characteristics of the system described by \(A\). Understanding how to derive and solve the characteristic equation is vital in analyzing any linear system.

Fundamental Set of Solutions

The concept of a "fundamental set of solutions" refers to a group of functions that form the basis for the solution space of a differential equation. For a system \(\mathbf{y}' = A\mathbf{y}\), solutions derived from eigenpairs are building blocks to express any general solution.

In our example, solutions were formulated using the eigenpairs; specifically, \(\mathbf{y}_1(t) = \begin{bmatrix}2 \ -3\end{bmatrix}\) for \(\lambda = 0\) and \(\mathbf{y}_2(t) = e^{-t}\begin{bmatrix}1 \ -\frac{9}{2}\end{bmatrix}\) for \(\lambda = -1\). To ensure these solutions form a fundamental set, they must be linearly independent. We check this by calculating their Wronskian, \(W(t)\).

If \(W(t) eq 0\) for any \(t\), the functions are independent, confirming they span the solution space's basis. Here, \(W(t) = -3e^{-t}\), which is nonzero for all \(t\). Therefore, the solutions indeed form a fundamental set, meaning any solution of the differential system can be represented by a linear combination of these solutions, crucial for describing the system's overall behavior.