If we have some system:
\begin{equation} x’ = Ax \end{equation}
the solution for this system should be \(e^{At}\). This gives rise to, given the power series:
\begin{equation} e^{At} = 1 + At + \frac{1}{2} \qty(At)^{2} + \frac{1}{3!} (At)^{3}+ \dots \end{equation}
the derivative of which:
\begin{align} \dv t e^{At} &= A + A^{2}t + \frac{A^{3}t^{2}}{2} + \dots \\ &= A\qty(1 + At + \frac{A^{2}t^{2}}{2}) \end{align}
This intuition makes sense for all matrices \(A\). Meaning the general solution gives:
\begin{equation} x = e^{At} x_0 \end{equation}
See also raising e to a matrix to see how to deal with diagonalizable matricies.
Benefits
- this approach produces all solutions no matter the eigenvalues of \(A\).
- also tells you what to do if your characteristic polynomial has repeated eigenvalues
- this is computationally not too bad if you have..
- diagonal \(A\)
- diagonalizable \(A\)
Great Matrix Exponential Tragedy
\begin{equation} e^{A+B} \neq e^{A} e^{B} \end{equation}
in general, because matricies don’t commute.