Problem 2 - Differential Equations and Linear Algebra

Question

Use the techniques from Section 9.4 and Section 9.5 to determine a fundamental matrix for \(\mathbf{x}^{\prime}=A \mathbf{x}\) and hence, find \(e^{A t}\). $$A=\left[\begin{array}{ll}2 & 1 \\\0 & 2\end{array}\right]$$.

Step-by-Step Solution

Answer

The matrix exponential \(e^{At}\) for the given matrix \( A=\left[\begin{array}{ll}2 & 1 \\0 & 2\end{array}\right]\) can be found by computing the fundamental matrix \(\Phi(t)\) using the eigenvalues and eigenvectors of A, and then calculating \(e^{At} = \Phi(t) \Phi^{-1}(0)\). The resulting matrix exponential is: \[e^{At} = \left[\begin{array}{cc}e^{2t}(t-1) & e^{2t} \\0 & e^{2t}\end{array}\right]\].

Step 1: Find the eigenvalues of the matrix A:

To determine the eigenvalues of A, first solve the characteristic equation: \[det(A - \lambda I) = 0\] Here, I is the identity matrix. Substitute A, and the equation becomes: \[\begin{vmatrix}2-\lambda & 1 \\0 & 2-\lambda \end{vmatrix} = 0\]

Step 2: Solve for the eigenvalues:

Now we can solve for the eigenvalues: \[(2-\lambda)^2 - (0\cdot1) = 0 \Rightarrow \lambda^2 - 4\lambda + 4 = 0\] By solving this quadratic equation, we find that the matrix A has only one eigenvalue: \[\lambda_1 = 2\]

Step 3: Find eigenvectors for the eigenvalue:

Now, we will find the eigenvectors corresponding to eigenvalue \(\lambda_1 = 2\). This involves solving the following linear system: \[(A - \lambda I)\mathbf{x} = 0\] Substitute \(\lambda_1\) and the equation becomes: \[\left[\begin{array}{ll}0 & 1 \\0 & 0\end{array}\right]\mathbf{x} = \left[\begin{array}{cc}0 \\ 0 \end{array}\right]\] Let \(\mathbf{x} = \left[\begin{array}{ll}x_{1} \\ x_{2} \end{array}\right]\). Then, the solutions for this equation are any scalar multiple of the vector \(\left[\begin{array}{cc}1 \\ 0 \end{array}\right]\).

Step 4: Find the fundamental matrix:

Since the matrix A has only one linearly independent eigenvector, it is called a defective matrix, and its fundamental matrix can be found using the following formula: \[\Phi(t) = \left[c_1 e^{\lambda_1t}(\mathbf{q}_1t + \mathbf{v}_1)\right]\] Here, \(\mathbf{q}_1\) is an eigenvector of A corresponding to \(\lambda_1\), and \(\mathbf{v}_1\) is a generalized eigenvector associated with \(\mathbf{q}_1\) and \(\lambda_1\). To find \(\mathbf{v}_1\), we will solve the following equation: \[(A - \lambda_1 I)\mathbf{v}_1 = \mathbf{q}_1\] Substitute \(\lambda_1\), \(\mathbf{q}_1 = \left[\begin{array}{cc}1 \\ 0 \end{array}\right]\), and the equation becomes: \[\left[\begin{array}{cc}0 & 1 \\0 & 0\end{array}\right]\mathbf{v}_1 = \left[\begin{array}{cc}1 \\ 0 \end{array}\right]\] Let \(\mathbf{v}_1 = \left[\begin{array}{cc}v_{11} \\ v_{12} \end{array}\right]\). Solving for \(\mathbf{v}_1\) gives us the vector \(\left[\begin{array}{cc}0 \\ 1 \end{array}\right]\). Now, we can find the fundamental matrix \(\Phi(t)\): \[\Phi(t) = \left[e^{2t} \left(\begin{array}{cc}t & 1\\\\0 & 1\end{array}\right)\right]\]

Step 5: Find the matrix exponential \(e^{At}\):

With the fundamental matrix \(\Phi(t)\), we can now find the matrix exponential \(e^{At}\) as follows: \[e^{At} = \Phi(t) \Phi^{-1}(0)\] Calculate the inverse of \(\Phi(0)\): \[\Phi(0) = \left[\begin{array}{cc}0 & 1 \\\\0 & 1\end{array}\right]\] \[\Phi^{-1}(0) = \frac{1}{(0 \cdot 1) - (1 \cdot 0)}\left[\begin{array}{cc}1 & -1 \\\\0 & 1\end{array}\right]\] Finally, calculate \(e^{At}\): \[e^{At} = \left[e^{2t} \left(\begin{array}{cc}t & 1\\\\0 & 1\end{array}\right)\right] \left[\begin{array}{cc}1 & -1 \\\\0 & 1\end{array}\right] = \left[\begin{array}{cc}e^{2t}(t-1) & e^{2t} \\\\0 & e^{2t}\end{array}\right]\] The matrix exponential \(e^{At}\) is given by: \[e^{At} = \left[\begin{array}{cc}e^{2t}(t-1) & e^{2t} \\\\0 & e^{2t}\end{array}\right]\]

Key Concepts

Eigenvalues

Eigenvalues are crucial when studying matrices in linear algebra. They are special scalars associated with a matrix, and they reveal many properties of the matrix. To find an eigenvalue of a matrix, we solve a specific equation called the characteristic equation. This equation is formed by subtracting \(\lambda I\) from the matrix \A\, where \I\ is the identity matrix, and then finding the determinant of the result and setting it equal to zero:

• Form the equation: \det(A - \lambda I) = 0\.
• Solve for \lambda\, which will give you the eigenvalues.

For the exercise matrix \(A = \begin{bmatrix}2 & 1 \0 & 2\end{bmatrix},\)there is only one eigenvalue \(\lambda = 2\), found by solving \[\det \left(\begin{array}{cc}2-\lambda & 1\0 & 2-\lambda\end{array}\right) = 0.\] This eigenvalue characterizes the effect of the matrix on vector directions in space.

Eigenvectors

Once we find an eigenvalue of a matrix, we need to find its corresponding eigenvectors. An eigenvector is a non-zero vector that changes only by a scalar factor when that corresponding matrix is applied. To find eigenvectors for a known eigenvalue, we solve the system formed by substituting the eigenvalue into:

• \( (A - \lambda I)\mathbf{x} = 0 \).

For example, for the eigenvalue \(\lambda = 2\) in our matrix:\[\left[\begin{array}{cc}0 & 1\0 & 0\end{array}\right]\left[\begin{array}{c}x_1\x_2\end{array}\right] = \left[\begin{array}{c}0\0\end{array}\right]\]This system results in eigenvectors that are scalar multiples of \([1, 0]^T\). Understanding eigenvectors helps in analyzing how matrices transform entire vector spaces.

Defective Matrix

In linear algebra, a defective matrix is one that does not have a complete set of linearly independent eigenvectors. This implies that it cannot be diagonalized. A matrix with repeated eigenvalues but fewer eigenvectors is classified as defective. Our matrix \( A = \begin{bmatrix} 2 & 1 \ 0 & 2 \end{bmatrix} \) is defective since it has only one independent eigenvector instead of two, for its eigenvalue \( \lambda = 2 \).

• To deal with a defective matrix, we use a generalized eigenvector.
• This helps in forming a fundamental matrix for solutions to differential equations.

It's essential to recognize defective matrices to know when additional steps, like finding generalized eigenvectors, are needed in calculations.

Fundamental Matrix

A fundamental matrix allows us to solve systems of first-order linear differential equations. It is constructed from a set of linearly independent solutions to the system, forming a matrix whose columns are each of these solutions. In this context, when dealing with a defective matrix, the construction of a fundamental matrix involves:

• Both an eigenvector and a generalized eigenvector.

For our exercise matrix, which is defective, the fundamental matrix \( \Phi(t) \) is given by:\[\Phi(t) = e^{2t} \begin{bmatrix} t & 1 \ 0 & 1 \end{bmatrix} \]This matrix captures the dynamics of the differential equation \( \mathbf{x}^{\prime} = A \mathbf{x} \).Ultimately, the fundamental matrix \( \Phi(t) \) can be used to calculate the matrix exponential \( e^{At} \), which is crucial for finding solutions to the differential equation.