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I start off with matrix $Q$, such that

$$Q = \begin{pmatrix} -\mu & \mu \\ \lambda & -\lambda \end{pmatrix}$$

and I want to work out the value of $P(t) = \exp(Qt)$

So to do this, you first diagonalise $Q$ and then work out the exponential of the diagonal matrix. I got this to be:

$${Q}t = \pmatrix{-\mu t &\mu t \\ \lambda t & -\lambda t} = \pmatrix{1 & -\frac{\mu }{\lambda } \\ 1 & 1} \cdot \pmatrix{0 & 0 \\ 0 & -t(\lambda + \mu)} \cdot \pmatrix{1 & -\frac{\mu }{\lambda } \\ 1 & 1}^{-1}.$$

Then, the next bit says

In this new basis, the infinitesimal generator is represented by the middle (diagonal) matrix and its exponential is

$$\exp{ \pmatrix{0 & 0 \\ 0 & -t(\lambda + \mu)}} = \pmatrix{\exp(0) & 0 \\ 0 & \exp(-t(\lambda + \mu))} = \pmatrix{1 & 0 \\ 0 & \exp(-t(\lambda + \mu))}.$$

Changing back to the original basis gives

$$\mathbb{P}(t) = \exp(\mathbb{Q} t) = \frac{1}{\lambda + \mu}\pmatrix{\lambda + \mu \exp(-t(\lambda+\mu)) & \mu - \mu \exp(-t(\lambda+\mu)) \\ \lambda - \lambda\exp(-t(\lambda+\mu)) & \mu + \lambda\exp(-t(\lambda+\mu))}.$$

How have they done this?

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1 Answer

up vote 4 down vote accepted

You can use that $e^{P^{-1}AP}=P^{-1}e^AP$

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@kaish this helped you? –  user52188 Jan 8 '13 at 21:08
    
Yeah it did and I was just going to complain and say that I can't get the right answer but looking at it again, I see that I tried to work out $Pe^AP^{-1}$ and not $P^{-1}e^AP$. I'll try that again and see if I get the right answer :) –  Kaish Jan 8 '13 at 21:55
    
great! note that $e^{PAP^{-1}}=Pe^AP^{-1}$ also –  user52188 Jan 8 '13 at 23:07
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