Why $(0.5,0.5)$ isn't stationary distribution for every 2 state ergodic Markov chain?

Question

Assume M is 2 state- ergodic Markov chain - i.e it's irreducible and aperiodic, with Stochastic matrix $Q \in \mathbb{R}^2 $ .
Then $Q(1,1) = (1,1)$. Hence $Q(\alpha,\alpha) = (\alpha,\alpha)$.
Now, since the stationary distribution (under these settings it exists and it's unique) holds that it's a probability, we can add the restriction that the sum of entries is 1 and get that $\alpha = 0.5$ .
Does it follows that $(0.5,0.5)$ is the stationary distribution of every ergodic 2 state Markov chain?

Answer 1

No. Take $$ Q=\pmatrix{\frac{2}{3}&\frac{1}{3}\\ \frac {3}{4}&\frac{1}{4}}\ , $$ for instance, for which the (unique) stationary distribution is $\ \pmatrix{\frac{9}{13}&\frac{4}{13}}\ .$ Your argument conflates multiplication on the right of $\ Q\ $ by a column vector with multiplication on its left by a row vector. In general, it is only the former for which $\ Q\pmatrix{1\\1}=\pmatrix{1\\1}\ ,$ while the stationary distribution. $\ \mu^T\ $ is only required to be invariant under the latter: $\ \mu^TQ=\mu^T\ .$