Exponential of $2\times 2$ matrix with first row equal to zero

Question

I would like to cumpute $\exp(tA)$ with $$A:=\begin{pmatrix}0&0\\\lambda&-\lambda\end{pmatrix}$$ and $0<t,\lambda$. If $A$ was diagonal, this would be simple, but I am stuck.

Motivation: Consider a $2$-level system (a switch) with two possible states $\bf 0$ (the switch is off) and $\bf 1$ (the switch is on). If the switch is on, it stays on for an exponential time and otherwise it stays off. In other words, the transition function is given by the following family of matrices: $$\forall t:\begin{pmatrix}p_t(\bf 0,\bf 0)&p_t(\bf 0,\bf 1)\\p_t(\bf 1,\bf 0)&p_t(\bf 1,\bf 1)\end{pmatrix}=\begin{pmatrix}1&0\\1-\exp(-\lambda t)&\exp(-\lambda t)\end{pmatrix}$$ We now define the $Q$-matrix as follows: $$Q:=\frac{\mathrm{d}}{\mathrm{d}t}p\big|_{t=0}=\begin{pmatrix}0&0\\\lambda&-\lambda\end{pmatrix}$$ As far as I know the transition function can be recovered from the $Q$-matrix through $\mathrm{e}^{tQ}=p_t$ and I would like to verify that.

Answer 1

Hint: look at $A^2, A^3, A^4$. Can you spot the pattern? Then substitute into $$ \exp(tA) = \sum_{n=0}^\infty t^n A^n/n!$$

Answer 2

I think that I was able to solve the problem using Robert's hint:

Suppose that $A$ is some matrix such that $A^2=\xi A$ for some number $\xi$. Then $A^n=\xi^{n-1}A$ for $n\geq 1$. In my case $\xi=-\lambda$ and we obtain \begin{equation} \exp(tA) = \sum_{n=0}^\infty t^n \frac{A^n}{n!}= I+\sum_{n=1}^\infty t^n \frac{A^n}{n!}=I+\sum_{n=1}^\infty t^n \frac{\xi^{n-1}A}{n!}=I+\left[\sum_{n=1}^\infty t^n \frac{\xi^n}{n!}\right]\frac{A}{\xi}\\=I+(\exp( t\xi)-1)\frac{A}{\xi}=I+(\exp(-\lambda t)-1)\frac{A}{-\lambda}=\begin{pmatrix}1&0\\0&1\end{pmatrix}+(\exp(-\lambda t)-1)\begin{pmatrix}0&0\\-1&1\end{pmatrix}=\begin{pmatrix}1&0\\1-\exp(-\lambda t)&\exp(-\lambda t)\end{pmatrix} \end{equation}