Transition Matrix of M/M/1 Queue We know that for an M/M/1 queue the state space is $S=\{0,1,2,... \}$. Further the probability to go from state $i$ to $i+1$ is $\lambda$ for all $i$ in $S$. Moreover, to go from $i$ to $i-1$ is the probability $\mu$ $\forall i \in S$.

So one can draw the following picture, taken from the wikipedia page on M/M/1 queues:



Now with this, one can establish the transition matrix:

$T=$\begin{pmatrix}
1-\lambda &amp; \lambda &amp; 0 &amp; 0 &amp; 0 &amp; ...\\
\mu &amp; 1-(\mu+ \lambda) &amp; \lambda &amp; 0 &amp; 0 &amp; ... \\
0 &amp; \mu &amp; 1-(\mu+\lambda) &amp; \lambda &amp; 0 &amp; ...\\
0 &amp; 0 &amp;  \mu &amp; 1-(\mu+\lambda) &amp; \lambda &amp; ...\\
&amp;&amp;&amp;&amp;&amp;...
\end{pmatrix}

I know that the transition rate matrix is given by:

$M=$\begin{pmatrix}
-\lambda &amp; \lambda &amp; 0 &amp; 0 &amp; 0 &amp; ...\\
\mu &amp; -(\mu+ \lambda) &amp; \lambda &amp; 0 &amp; 0 &amp; ... \\
0 &amp; \mu &amp; -(\mu+\lambda) &amp; \lambda &amp; 0 &amp; ...\\
0 &amp; 0 &amp;  \mu &amp; -(\mu+\lambda) &amp; \lambda &amp; ...\\
&amp;&amp;&amp;&amp;&amp;...
\end{pmatrix}

How can I get from the transition matrix to the transition rate matrix? Is it even possible?

I appreciate any help! Thank you very much!
More generally... one may want to keep in mind that Markov chains (in discrete time) and Markov processes (in continuous time) are different (although related) objects.

1. A Markov chain $(X_n)$, indexed by $n$ integer, is described by a transition matrix $P$, such that, for every states $(x,y)$ and every $n$, $$\Pr(X_{n+1}=y\mid X_n=x)=P_{xy}.$$ Thus: 


  The matrix $P$ has nonnegative entries and every row sum $\sum\limits_yP_{xy}$ is $1$. 


The powers of the matrix $P$ yield the transition probabilities of the chain in any number of steps, through the identities 


  $$\Pr(X_{n+k}=y\mid X_n=x)=(P^k)_{xy}.$$


2. A Markov process $(Y_t)$, indexed by $t$ real, is described by a transition rate matrix $Q$, often called its generator, such that, informally, for every states $x\ne y$, when $\mathrm dt\to0^+$, $$\Pr(X_{t+\mathrm dt}=y\mid X_t=x)=Q_{xy}\mathrm dt+o(\mathrm dt).$$ Additionally, the diagonal entries are defined as $$Q_{xx}=-\sum_yQ_{xy}.$$
Thus:


  The offdiagonal entries of the matrix $Q$ are nonnegative and every row sum $\sum\limits_yQ_{xy}$ is $0$. 


The matrix $Q$ yields the transition probabilities of the process, through the identities 


  $$\Pr(X_{t+s}=y\mid X_t=x)=(\mathrm e^{sQ})_{xy},$$ 


for every nonnegative $(t,s)$, where the exponential is defined by the usual series, always convergent, that is, $$\mathrm e^{sQ}=\sum_{n\geqslant0}\frac{s^n}{n!}Q^n.$$

3. Every Markov process gives rise to some Markov chains, usually called embedded chains, by discretizations of the time. That is, if $(Y_t)$ is a Markov process with generator $Q$ , then for each $\tau\gt0$, $(X_n)$ defined by $$X_n=Y_{n\tau},$$ is a Markov chain with transition matrix $$P=\mathrm e^{\tau Q}.$$ (Exercise: Show that if $Q$ is a generator, that is, with nonnegative offdiagonal entries and every row sum equal to $0$, then $\mathrm e^{\tau Q}$ is a transition matrix, that is, with nonnegative entries and every row sum equal to $1$.)

4. In the other direction, not every Markov chain corresponds to the discretization of a Markov process. The ones that do are called embeddable. 

(Exercise: Find some non embeddable transition matrix $P$. Hint: Two states are enough, and watch out for the zero entries of $P$.)

5. Queues in general, and M/M/1 queues in particular, are Markov processes (in continuous time), as explained on the WP page the OP links to. As such, they can be discretized into Markov chains (in discrete time).

Nota: All this can be made rigorous on every finite state space, always, and on countable state spaces, under some technical additional conditions that we left out (and that are satisfied for the queues the OP is interested in).
The matrix is wrong. It should read "1-$\lambda$" instead

Also for the other probabilities it is not $-(\mu + \lambda)$ but rather $1-(\mu + \lambda)$
as probabilities can't be negative, also it's pretty clear by the picture you posted.