Question

find a fundamental matrix for the given system of equations. In each case also find the fundamental matrix \(\mathbf{\Phi}(t)\) satisfying \(\Phi(0)=\mathbf{1}\) $$ \mathbf{x}^{\prime}=\left(\begin{array}{rrr}{1} & {1} & {1} \\ {2} & {1} & {-1} \\ {-8} & {-5} & {-3}\end{array}\right) \mathbf{x} $$

Short answer

Based on the given information, determine the fundamental matrix \(\mathbf{\Phi}(t)\), where \(\mathbf{\Phi}(0) = \mathbf{1}\). Solution: The fundamental matrix \(\mathbf{\Phi}(t)\) is given by: $$\mathbf{\Phi}(t)=\frac{1}{2}\left(\begin{array}{ccc}{1} & {1} & {1} \\\ {1} & {-1} & {1} \\\ {-2} & {1} & {-1}\end{array}\right)\left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right) \left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$

Step-by-step solution

Determine the eigenvalues and eigenvectors of the given matrix

To find eigenvalues and their respective eigenvectors, we need to solve the following equation: $$\text{det} \ |A - \lambda I| = 0$$ where \(A\) is the given matrix and \(\lambda\) represents eigenvalues. The given matrix is: $$A=\left(\begin{array}{rrr}{1} & {1} & {1} \\\ {2} & {1} & {-1} \\\ {-8} & {-5} & {-3}\end{array}\right)$$ The equation becomes: $$\text{det} \ \left(\begin{array}{ccc}{1-\lambda} & {1} & {1} \\\ {2} & {1-\lambda} & {-1} \\\ {-8} & {-5} & {-3-\lambda}\end{array}\right) = 0$$ Solve the determinant to find the eigenvalues. After solving the determinant, we get the eigenvalues: \(\lambda_1 = -1\), \(\lambda_2 = 1\), and \(\lambda_3 = -3\). Now, find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = -1\): $$\left(\begin{array}{ccc}{2} & {1} & {1} \\\ {2} & {2} & {-1} \\\ {-8} & {-5} & {-2}\end{array}\right)\mathbf{x} = \mathbf{0}$$ The eigenvector is: \(\mathbf{x}_1 = (1,1,-2)^T\) For \(\lambda_2 = 1\): $$\left(\begin{array}{ccc}{0} & {1} & {1} \\\ {2} & {0} & {-1} \\\ {-8} & {-5} & {-4}\end{array}\right)\mathbf{x} = \mathbf{0}$$ The eigenvector is: \(\mathbf{x}_2 = (1,-1,1)^T\) For \(\lambda_3 = -3\): $$\left(\begin{array}{ccc}{4} & {1} & {1} \\\ {2} & {4} & {-1} \\\ {-8} & {-5} & {0}\end{array}\right)\mathbf{x} = \mathbf{0}$$ The eigenvector is: \(\mathbf{x}_3 = (1,1,-1)^T\)

Find the matrix exponential of the given matrix

Using the eigenvectors, we can form a matrix \(X=[\mathbf{x}_1,\mathbf{x}_2,\mathbf{x}_3]\) which contains eigenvectors as its columns. We then find a diagonal matrix \(\Lambda\) with the eigenvalues as its elements along the diagonal, that is, $$X=\left(\begin{array}{ccc}1 & 1 & 1 \\\ 1 & -1 & 1 \\\ -2 & 1 & -1\end{array}\right), \quad \Lambda=\left(\begin{array}{ccc}-1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & -3\end{array}\right)$$ Now, we can compute the matrix exponential of \(A\) as follows, $$\Phi(t) = e^{A t} = X e^{\Lambda t}X^{-1}$$ where the exponential of a diagonal matrix is calculated by taking the exponential of all the diagonal elements separately. Thus we get, $$e^{\Lambda t} = \left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right)$$ Now, we need to calculate \(X^{-1}\), which is the inverse of matrix \(X\). $$X^{-1} = \frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$ Finally, calculate \(\Phi(t)\): $$\Phi(t) = Xe^{\Lambda t}X^{-1} = \left(\begin{array}{ccc}1 & 1 & 1 \\\ 1 & -1 & 1 \\\ -2 & 1 & -1\end{array}\right)\left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right) \frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$

Find the desired fundamental matrix satisfying \(\Phi(0)=\mathbf{1}\)

The fundamental matrix fulfilling this condition can be obtained by evaluating \(\Phi(t)\) at \(t=0\) and solving for the constant matrix \(C\). Then multiply \(\Phi(t)\) by the constant matrix \(C\). First, calculate \(\Phi(0)\): $$\Phi(0) = \left(\begin{array}{ccc}1 & 1 & 1 \\\ 1 & -1 & 1 \\\ -2 & 1 & -1\end{array}\right)\left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right) \frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$ $$\Phi(0) = \left(\begin{array}{ccc}{1} & {1} & {1} \\\ {1} & {-1} & {1} \\\ {-2} & {1} & {-1}\end{array}\right)\frac{1}{2}\left(\begin{array}{rrr}{-1} & {1} & {1} \\\ {2} & {0} & {1} \\\ {-1} & {-1} & {0}\end{array}\right)$$ $$\Phi(0) = \left(\begin{array}{ccc}{0} & {0} & {2} \\\ {0} & {2} & {0} \\\ {0} & {0} & {2}\end{array}\right)$$ Now, to find the constant matrix \(C\), we need \(\Phi(0)\) to be equal to the identity matrix: $$\Phi(0)C = \mathbf{1}$$ $$\left(\begin{array}{ccc}{0} & {0} & {2} \\\ {0} & {2} & {0} \\\ {0} & {0} & {2}\end{array}\right)C = \left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$ Solve for \(C\): $$C = \frac{1}{2}\left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$ Finally, compute the desired fundamental matrix as \(\Phi(t)C\): $$\mathbf{\Phi}(t)=\Phi(t)C=\frac{1}{2}\left(\begin{array}{ccc}{1} & {1} & {1} \\\ {1} & {-1} & {1} \\\ {-2} & {1} & {-1}\end{array}\right)\left(\begin{array}{ccc}{e^{-t}} & {0} & {0} \\\ {0} & {e^t} & {0} \\\ {0} & {0} & {e^{-3t}}\end{array}\right) \left(\begin{array}{ccc}{1} & {0} & {0} \\\ {0} & {1} & {0} \\\ {0} & {0} & {1}\end{array}\right)$$

Key concepts

Eigenvalues and Eigenvectors

Let's dive into the fascinating world of eigenvalues and eigenvectors. These concepts play a crucial role in understanding linear transformations and systems of differential equations.
To find the eigenvalues of a matrix, we need to solve the equation \( \text{det}(A - \lambda I) = 0 \). This determinant equation helps us find points where the transformation described by the matrix "stretches" or "shrinks" the space by a constant factor \( \lambda \), known as the eigenvalue.

Once we find the eigenvalues, we can determine the eigenvectors. An eigenvector is a non-zero vector that remains parallel to itself after being transformed by the matrix. In other words, when we apply the matrix to this vector, the direction is preserved even if the magnitude changes by the eigenvalue.
Here’s a quick breakdown:

• Eigenvalues are special numbers associated with a matrix that provide insights into the matrix’s properties.
• Eigenvectors are vectors that maintain their direction under the transformation depicted by the matrix.

With \( \lambda_1 = -1 \), \( \lambda_2 = 1 \), and \( \lambda_3 = -3 \) for our matrix, we find corresponding eigenvectors like \( (1, 1, -2)^T \), which reveal how the matrix maps space along particular directions.

Matrix Exponential

Matrix exponentials are a captivating concept that helps us solve systems of linear differential equations, particularly when dealing with stability and dynamics.
The exponential of a matrix \( A \) is a powerful tool defined as \( e^{At} \). It's similar to the exponential function in real numbers but adapted for matrices. Calculating the matrix exponential can be complex, but when a matrix has distinct eigenvalues, it becomes more manageable.

Utilizing the matrix of eigenvectors \( X \) and the diagonal matrix \( \Lambda \), the exponential of \( A \) for our problem can be found with the formula:

• \( e^{At} = X e^{\Lambda t} X^{-1} \)
• \( e^{\Lambda t} \) is straightforward as we raise the exponential to the power of each eigenvalue along the diagonal of \( \Lambda \)

This approach allows us to break down a seemingly daunting calculation into smaller, more manageable parts. The simplicity of \( e^{\Lambda t} \), with exponentials acting on the eigenvalues, provides a foundation for computing \( e^{At} \). This matrix exponential subsequently provides a solution to the system of differential equations involved.

System of Differential Equations

Solving a system of differential equations can seem overwhelming, but breaking it down into steps makes it more approachable. Systems of differential equations model many natural phenomena, from population dynamics to electrical circuits. The foundation often lies in linear differential equations, such as the one in our exercise.
Here's a glance at how it works:

• First, we represent the system with a matrix equation \( \mathbf{x}' = A\mathbf{x} \), where \( \mathbf{x} \) is a vector of functions.
• The goal is to find a matrix \( \Phi(t) \), known as the fundamental matrix, that describes how solutions to the system behave over time.

Finding \( \Phi(t) \) requires understanding the matrix's eigenvalues and eigenvectors. The calculation of \( e^{At} \), which uses these values, gives us the solution to the system by mapping initial conditions through time. Solving for \( \Phi(t) \) can be seen as finding a bridge between initial states and their evolution, making it a powerful technique for studying complex systems.
By understanding these concepts, we grasp how systems change and interact over time, providing insights into their long-term behavior.