Does the stationary distribution of this Markov Chain exist?

To find the stationary distribution of a Markov Chain, I believe I must solve for $\vec{s} = \langle s_0, s_1 \rangle$ in $\vec{s} = \vec{s}Q$, where $Q$ is the transition matrix.

$Q$, in my case, is

$$\left( \begin{array}{cc} p & 1-p \\ 1-q & q \end{array} \right)$$

where $Q_{ij}$ is the probability of moving from state $i$ to state $j$ (row $i$, column $j$). When I solve for $s_0$ and $s_1$, however, I get

$$s_0 = s_0 p + s_1 (1-q) \\ s_1 = s_0 (1 - p) + s_1 q$$

Subsequently,

$$s_0 (1 - p) = s_1 (1 - q) \\ s_1 (1 - q) = s_0 (1 - p)$$

These two equations look identical. Does that mean there are an infinite number of stationary distributions for this Markov chain?

Thanks for helping a Markov Chain newb :)

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no, it does not. Remember that you have an additional condition $s_0+s_1=1$. –  Artem Dec 8 '12 at 1:35
Ah, thank you. Why does that have to be a condition? –  David Faux Dec 8 '12 at 1:38
Because these are probabilities which have to sum to one. Or I don't understand your question? –  Artem Dec 8 '12 at 1:38
Ohhh... wait, why must the probabilities sum to 1? –  David Faux Dec 8 '12 at 1:41
Well, think what it would mean if probabilities added up to $3$ --- or to $-7$. –  Gerry Myerson Dec 8 '12 at 6:00

As in the comments, the condition you are missing is that the probabilities must sum to 1 ($s_0+s_1=1$).
It's a property of the $Q$ matrix. The $Q$ matrix has rank 1 less than its dimension (here rank 1) because the final column is redundant. If it were left blank you could 'fill in the gaps': $$\begin{pmatrix} p & *\\ 1-q & ** \end{pmatrix}.$$ You know that each row is a probability vector because it describes what can potentially happen. When in state $0$ the chain can remain in state $0$ or move to state $1$. As these are the only things that can happen the probabilities $p+*=1$ so $*=1-p$. Similarly $**=q$.