Houjun Liu

matrix exponentiation

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.


  1. this approach produces all solutions no matter the eigenvalues of \(A\).
  2. also tells you what to do if your characteristic polynomial has repeated eigenvalues
  3. this is computationally not too bad if you have..
    1. diagonal \(A\)
    2. 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.