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I' am reading Introduction to Stochastic Processes by Lawler and I' am a bit confused how demonstrates you get the transition matrix $\textbf{P}_t$ from the infinitesimal generator $\textbf{A}$. I'll provide the example in the book:

$\textbf{Example 1.}$ Consider a chain with two states, (0,1). Assume $\alpha(0,1)=1$ and $\alpha(1,0)=2$. Then the infinitesimal generator is

$$\textbf{A}=\left[\begin{array}{cc} -1 & 1 \\ 2 & -2 \end{array}\right]$$

In order to compute $e^{t\textbf{A}}$, we diagonalize the matrix. The eigenvalues are 0, -3. We can write

$$\textbf{D}=\textbf{Q}^{-1}\textbf{A}\textbf{Q}$$

where

$$\textbf{D}=\left[\begin{array}{cc} 0 & 0 \\ 0 & -3 \end{array}\right],\textbf{Q}=\left[\begin{array}{cc} 1 & 1 \\ 1 & -2 \end{array}\right], \textbf{Q}^{-1}=\left[\begin{array}{cc} 2/3 & 1/3 \\ 1/3 & -1/3 \end{array}\right]$$

We use the diagonalization to compute the exponential $e^{t\textbf{A}}$.

\begin{align} \textbf{P}_t&=e^{t\textbf{A}}=\sum_{n=0}^{\infty}\frac{(t\textbf{A})^n}{n!} \\ &=\sum_{n=0}^{\infty}\frac{\textbf{Q}(t\textbf{D})^n\textbf{Q}^{-1}}{n!} \\ &=\textbf{Q}\left[\begin{array}{cc} 1 & 0 \\ 0 & e^{-3t} \end{array}\right]\textbf{Q}^{-1} \\ &=\left[\begin{array}{cc} 2/3 & 1/3 \\ 2/3 & 1/3 \end{array}\right] + e^{-3t}\left[\begin{array}{cc} 1/3 & -1/3 \\ -2/3 & 2/3 \end{array}\right] \end{align}

What I don't understand is what the second to last line is not,

$$\textbf{Q}\left[\begin{array}{cc} 1 & 1 \\ 1 & e^{-3t} \end{array}\right]\textbf{Q}^{-1} $$

Though I tried it out and it would not provide a valid stationary distribution.

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I just didn't understand exponentiation of a matrix. Reading Wikipedia (don't have my linear algebra book with me) I see the book is correct because exponentiating a diagonal matrix yields all zeros on the off-diagonal: https://en.wikipedia.org/wiki/Matrix_exponential#Diagonalizable_case

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    $\begingroup$ Yes. This is because the exponential comes from summing $(tD)^n/n!$, and the diagonal power $D^n$ is simply the diagonals raised to the power of $n$. You can accept your own answer ;) $\endgroup$ – P.Windridge Feb 13 '16 at 22:33

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