Why doesn't detailed balance always imply a uniform distribution for Markov Chains?

Question

Consider a state (row) vector $\{\pi_i\}$ and a right-stochastic matrix $P_{ij}$. To calculate the next state vector $\pi'$ you can right-multiply by $P$: $$\pi'_i=\sum_j \pi_jP_{ji}$$ Detailed balance states that in equilibrium $$\pi_iP_{ij}=\pi_jP_{ji}.$$ If you sum over both sides you get the following \begin{align} \sum_i\pi_iP_{ij}&=\sum_j\pi_jP_{ji}\\ \pi'_j&=\pi'_i \end{align} which seems to imply that for any $P_{ij}$ obeying the detailed balance equation, the stationary distribution is the uniform distribution since every entry of $\pi'$ is the same. However in this article it is mentioned that only when $P$ is symmetric do we always get a uniform distribution. How can this be? I also don't fully understand the detailed balance so this might add to the confusion

Answer 1

In response to @nejimban: In the step where I introduced the sums I had the following idea in mind, starting from a detailed balance equation for a particular index just keep adding terms from different indices until you have a full sum. Writing this down explicitly made me realize this is not possible. I might have used Einstein-summation too much \begin{align} \pi_1P_{13}&=\pi_3P_{31}\\ \pi_1P_{13}+\color{blue}{\pi_2P_{23}}&=\pi_3P_{31}+\color{blue}{\pi_2P_{23}}\\ \pi_1P_{13}+\pi_2P_{23}+\color{blue}{\pi_3P_{33}}&=\pi_3P_{31}+\pi_2P_{23}+\color{blue}{\pi_3P_{33}}\\ \pi'_3&=\pi_3P_{31}+\pi_2P_{23}+\color{}{\pi_3P_{33}}\\ \pi'_3&\neq\pi'_1 \end{align}