# Exponential of $2\times2$ matrix with double eigenvalue $-1$

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.

Thanks.

• Hint: $A = -I + N$ where $N^2 = 0$. Compute $A^n$ and use the usual series for the exponential. – Gribouillis Aug 25 '17 at 16:17
• @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 $J_2$? – MathIsHard Aug 25 '17 at 16:20
• You can also use the Cayley-Hamilton theorem: en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem – md2perpe Aug 25 '17 at 16:23

## 2 Answers

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.

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

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