(I) Fundamental Matrix Solution

Consider the homogeneous linear system \(\vec{x}' = \mathbf{A}\vec{x}\). Let \(\vec{x}_1(t), \dots, \vec{x}_n(t)\) be \(n\) linearly independent solutions. A **fundamental matrix** \(\Phi(t)\) is constructed by placing these solutions as its columns:

\[ \Phi(t) = \begin{bmatrix} \vec{x}_1(t) & \dots & \vec{x}_n(t) \end{bmatrix} \]

Relationship Between Fundamental Matrices

A fundamental matrix is not unique. If \(\Psi(t)\) is any other fundamental matrix for the same system, then each column of \(\Psi(t)\) is a linear combination of the columns of \(\Phi(t)\). Therefore, there exists a constant \(n \times n\) non-singular matrix \(\mathbf{M}\) such that:

\[ \Psi(t) = \Phi(t)\mathbf{M} \]

Theorem 1: Fundamental Matrix Solutions

The unique solution to the Initial Value Problem \(\vec{x}' = \mathbf{A}\vec{x}, \vec{x}(0) = \vec{x}_0\) is given by:

\[ \vec{x}(t) = \Phi(t)\Phi(0)^{-1}\vec{x}_0 \]
Example 1

Solve: \( x' = 4x + 2y, y' = 3x - y \) with \( x(0) = 1, y(0) = -1 \).

Step 1: Fundamental Matrix. Two solutions are \(\vec{x}_1 = \begin{bmatrix} e^{-2t} \\ -3e^{-2t} \end{bmatrix}, \vec{x}_2 = \begin{bmatrix} 2e^{5t} \\ e^{5t} \end{bmatrix}\). Thus:

\[ \Phi(t) = \begin{bmatrix} e^{-2t} & 2e^{5t} \\ -3e^{-2t} & e^{5t} \end{bmatrix} \implies \Phi(0) = \begin{bmatrix} 1 & 2 \\ -3 & 1 \end{bmatrix} \implies \Phi(0)^{-1} = \frac{1}{7} \begin{bmatrix} 1 & -2 \\ 3 & 1 \end{bmatrix} \]

Step 2: Solution. Using \(\vec{x}(t) = \Phi(t)\Phi(0)^{-1}\vec{x}_0\):

\[ \vec{x}(t) = \begin{bmatrix} e^{-2t} & 2e^{5t} \\ -3e^{-2t} & e^{5t} \end{bmatrix} \cdot \frac{1}{7} \begin{bmatrix} 1 & -2 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} 3e^{-2t} + 4e^{5t} \\ -9e^{-2t} + 2e^{5t} \end{bmatrix} \]

(II) Exponential Matrices

Matrix exponentials are defined by the series:

\[ e^{\mathbf{A}t} = \mathbf{I} + \mathbf{A}t + \frac{\mathbf{A}^2 t^2}{2!} + \dots + \frac{\mathbf{A}^n t^n}{n!} + \dots \]
Example 2: Nilpotent Decomposition

Let \(\mathbf{A} = \begin{bmatrix} 2 & 3 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 2 \end{bmatrix}\). Decompose into \(\mathbf{A} = 2\mathbf{I} + \mathbf{B}\) where \(\mathbf{B}\) is nilpotent.

Step 1: Identity Part. The exponential of the identity multiple is:

\[ e^{2\mathbf{I}t} = \begin{bmatrix} e^{2t} & 0 & 0 \\ 0 & e^{2t} & 0 \\ 0 & 0 & e^{2t} \end{bmatrix} \]

Step 2: Nilpotent Part. Since \(\mathbf{B}^2 = \begin{bmatrix} 0 & 0 & 18 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\) and \(\mathbf{B}^3 = \mathbf{0}\):

\[ e^{\mathbf{B}t} = \mathbf{I} + \mathbf{B}t + \frac{1}{2}\mathbf{B}^2t^2 = \begin{bmatrix} 1 & 3t & 4t + 9t^2 \\ 0 & 1 & 6t \\ 0 & 0 & 1 \end{bmatrix} \]

Step 3: Final Matrix. Since \(2\mathbf{I}t\) and \(\mathbf{B}t\) commute:

\[ e^{\mathbf{A}t} = e^{2\mathbf{I}t}e^{\mathbf{B}t} = \begin{bmatrix} e^{2t} & 3te^{2t} & (4t+9t^2)e^{2t} \\ 0 & e^{2t} & 6te^{2t} \\ 0 & 0 & e^{2t} \end{bmatrix} \]

(III) Matrix Exponential Solutions

Theorem 2: Exponential Solutions

The unique solution to \(\vec{x}' = \mathbf{A}\vec{x}, \vec{x}(0) = \vec{x}_0\) is:

\[ \vec{x}(t) = e^{\mathbf{A}t}\vec{x}_0 \]

This implies that

\[ e^{\mathbf{A}t} = \Phi(t)\Phi(0)^{-1} \]

Example 3: Constructing eAt

For \(\mathbf{A} = \begin{bmatrix} 4 & 2 \\ 3 & -1 \end{bmatrix}\), using \(\Phi(t)\) from Example 1:

\[ e^{\mathbf{A}t} = \frac{1}{7} \begin{bmatrix} e^{-2t} & 2e^{5t} \\ -3e^{-2t} & e^{5t} \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 3 & 1 \end{bmatrix} = \frac{1}{7} \begin{bmatrix} e^{-2t} + 6e^{5t} & -2e^{-2t} + 2e^{5t} \\ -3e^{-2t} + 3e^{5t} & 6e^{-2t} + e^{5t} \end{bmatrix} \]
Example 4: Solving IVP

Solve \(\vec{x}' = \begin{bmatrix} 2 & 5 \\ 0 & 2 \end{bmatrix}\vec{x}, \vec{x}(0) = \begin{bmatrix} 4 \\ 7 \end{bmatrix}\).

Decompose \(\mathbf{A} = 2\mathbf{I} + \mathbf{B}\). Then \(e^{\mathbf{A}t} = e^{2\mathbf{I}t}e^{\mathbf{B}t}\):

\[ e^{\mathbf{A}t} = \begin{bmatrix} e^{2t} & 0 \\ 0 & e^{2t} \end{bmatrix} \begin{bmatrix} 1 & 5t \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} e^{2t} & 5te^{2t} \\ 0 & e^{2t} \end{bmatrix} \]

Final Particular Solution:

\[ \vec{x}(t) = \begin{bmatrix} e^{2t} & 5te^{2t} \\ 0 & e^{2t} \end{bmatrix} \begin{bmatrix} 4 \\ 7 \end{bmatrix} = \begin{bmatrix} (4 + 35t)e^{2t} \\ 7e^{2t} \end{bmatrix} \]