I am trying to learn how to do the exponential of a matrix $A= \begin{bmatrix} 0 & 1 \\ -1 & -2 \end{bmatrix}$

I found the eigenvalues to be $-1$ and $-1$. The eigenvector is $(1,-1)$ and the generalized eigenvector $(1,0)$.

The jordan form of the matrix is $\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$

I am not sure what to do now to solve $e^{At}$. Any help would be much appreciated. I have been coming back to this problem for a few days now.


  • $\begingroup$ Hint: $A = -I + N$ where $N^2 = 0$. Compute $A^n$ and use the usual series for the exponential. $\endgroup$ – Gribouillis Aug 25 '17 at 16:17
  • $\begingroup$ @Moo I actually have been looking at that. I don't understand the last steps where it says $e^{At} = Te^{J_2(-t)}T^{-1}$. Why $e^{J_2(-t)}$? and What is $J_2$? $\endgroup$ – MathIsHard Aug 25 '17 at 16:20
  • $\begingroup$ You can also use the Cayley-Hamilton theorem: en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem $\endgroup$ – md2perpe Aug 25 '17 at 16:23

For the given example the hint of Gribouillis is perhaps the easiest approach. Here some extra hints: Since $IN=NI=N$ (in particular they commute) you have: $$ \exp(tA) = \exp(t(-I+N)) = \exp(-tI) \exp(tN)=e^{-t} \exp(tN)$$ Now for the term $\exp(tN)$ use the Taylor expansion, and the result, due to $N^2=0$, is really simple...


$A = PJP^{-1}$

I assume you can find the generalized eigenvectors

$e^{At} = \sum \frac {(At)^n}{n!} = P (\sum \frac {(Jt)^n}{n!})P^{-1}$

$\begin{bmatrix} -1&1\\0&-1\end{bmatrix}^n = \begin{bmatrix} (-1)^n &(-1)^{n-1}n\\0&(-1)^n\end{bmatrix}$

If this is not obvious, you can prove it with induction.

$\sum \frac {(-1)^{n}t^n}{n!} = e^{-t}\\ \sum \frac {n(-1)^{n-1}t^n}{n!} = \sum \frac {(-1)^{n-1}t^n}{(n-1)!} = t\sum \frac {(-1)^nt^n}{(n)!} = te^{-t}$

$e^{At} = P \begin{bmatrix} e^{-t} & te^{-t}\\0&e^{-t}\end{bmatrix}P^{-1}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.