Diagonalization of a Matrix. Having Trouble..

Question

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.

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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.

Answer 1

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.