This is for Continuous-time Markov chains but I'm having trouble with the linear algebra.
Formal Definitions:
If Q is diagonalizable, then so is $e^{tQ}$, and the transition function can be expressed in terms of the eigenvalues and eigenvectors of $Q$. Write $Q=SDS^{-1}$, where $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $Q$, and $S$ is an invertible matrix whose columns are the corresponding eigenvectors. This gives,
$$e^{tQ} = Se^{tD}S^{-1}$$
The Problem:
A Markov Chain has generator matrix,
$$Q= \begin{pmatrix} -1 & 1 & 0 \\ 0 & -2 & 2 \\ 3 & 0 & -3 \\ \end{pmatrix} $$
Find the transition function by diagonalizing the generator and finding the matrix exponential.
My pr0fessor seems to be getting eigenvalues $\lambda = -4, \lambda = -2, \lambda = 0$.
I, on the other hand, seem to be getting $\lambda = -3, \lambda = 0$.
Edit: I recalculated but still am getting $\lambda = 0$.
This gets him a completely different set of eigenvectors.. Am I miscalculating something?
EDIT:
In the back of the book I have,
$$P(t)= \begin{pmatrix} -1 & 0 & 1 \\ -3 & 1 & 1 \\ 3 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} e^{-4t} & 0 & 0 \\ 0 & e^{-2t} & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} -1/4 & 0 & 1/4 \\ -3/2 & 1 & 1/2 \\ 3/4 & 0 & 1/4 \\ \end{pmatrix} $$
Where $P(t)$ is the transition function asked for.
There's a typo. Reverse engineering the given answer we have: $$ \begin{align} Q&=PDP^{-1}\\ &=\begin{pmatrix} -1 & 0 & 1 \\ -3 & 1 & 1 \\ 3 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} -4 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} -1/4 & 0 & 1/4 \\ -3/2 & 1 & 1/2 \\ 3/4 & 0 & 1/4 \\ \end{pmatrix}\\ &=\begin{pmatrix} -1 & 0 & 1 \\ 0 & -2 & 2 \\ 3 & 0 & -3 \\ \end{pmatrix} \end{align} $$
Now knowing that $$ Q=\begin{pmatrix} -1 & 0 & 1 \\ 0 & -2 & 2 \\ 3 & 0 & -3 \\ \end{pmatrix} $$ You can find your characteristic polynomial, e.g. via cofactor expansion down the second column to get $$p(\lambda)=-\lambda(\lambda+2)(\lambda+4)$$ and proceed from there.