Stationary distribution of convex combination of Markov chains Let $P$ be a stochastic matrix (of an irreducible Markov Chain) with stationary distribution $\pi^T$ (i.e. $\pi^T P = \pi^T$) and let further $E$ be the matrix of all $1$'s. 

Given an $\alpha \in [0,1]$, is it possible to find an expression for the stationary distribution of 
$$\alpha P + \frac{(1-\alpha)}{n}E,$$
depending on $\pi$ and $\frac{1}{n}\mathbb{1}$, where $\mathbb{1}$ is the vector of all $1$'s?

More generally; given two transition matrices of irreducible Markov Chains $P_1$ and $P_2$ with stationary distributions $\pi_1^T$ and $\pi_2^T$, respectively. Can one find a general formula to calculate the stationary distribution of 
$$\alpha P_1 + (1-\alpha)P_2 \quad,$$
for $\alpha \in [0,1]$?
 A: If all you know is the stationary distributions, then there is no general formula.
Here is a simple example:
Take $P$, $Q$ to be irreducible, and $A$ the everywhere self loop chain, then $P := eP + (1-e) A$ and $Q' := fQ + (1-f) A$ are irreducible with the same stationary distributions as $P$ and $Q$. $P'$ and $Q'$ are basically just lazy versions of $P$ and $Q$. 
Moreover, for every fixed $\lambda \in (0,1)$, you can pick different $e,f \in (0,1)$ to make $ X_{\lambda}(e,f) := \lambda P' + (1 - \lambda) Q' = \lambda e P + \lambda f Q + (1 - \lambda)( 2 - e - f) A$ have extremely different stationary distributions.
Specifically, you can show that for all $\lambda$, as the stationary distribution of $X_{\lambda}(e,f)$ converges to $\pi_P = \pi_{P'}$ as $f \to 0$ and it converges to $\pi_Q = \pi_{Q'}$ as $e \to 0$.

This is a lot of words -- all I'm saying is something very intuitive: if $P'$ is extremely lazy, then mixing it with $Q'$, may have very little effect on the stationary distribution of the mixture. Like if $P'$ is so lazy that it only takes a step to a new state once in a million years, and $Q'$ energetically takes steps to new states every second, then alternating $P'$ and $Q'$ is basically indistinguishable from a lazier version of $Q'$.

Note that you can talk about continuity of the stationary distribution in $\lambda$ -- by considering the $1$ eigenspaces of $P$ and $Q$, this is basically saying that if I have a continuous path of matrices, with one dimensional kernels, then the kernels change continuously. (This is how I would show the claim about letting $e \to 0$ above.)
Proof: If $A_n \to A$, let $v_n$ generate the kernels of $A_n$. We can assume $||v_n|| = 1$, so by compactness we can pass to a convergent subsequence, say $v_n \to v$. 
Then we have $A_n(v_n) = ( A_n(v_n) - A(v_n) ) + (A(v_n) - A(v)) + A(v)$.
$||A_n(v_n) - A(v_n) ||_2 \leq ||A_n - A||_2 \to 0$, and $A(v_n) - A(v) \to 0$ as $n$ to infinity. So we get that $A_n(v_n) \to A(v)$ by triangle inequality. As $A_n(v_n) = 0$ we obtain that $A(v) = 0$. As $||v_n|| = 1$, we get $||v|| = 1$.
Since $A$ has one dimensional kernel, it is generated by $v$. (Note that in general the kernel of $A$ could jump in dimension -- this might correspond to the case of $P$ have a unique stationary distribution, but $Q$ not having a unique stationary, e.g. not being ergodic.)

NB: The relationship between $A$ and $ker(A)$ is actually algebraic: By using $ker(A) = im(A^T)^{\perp}$ (an isomorphic isometry on the relevant Grassmannians), it suffices to control the rate of change of the image. For a given matrix $A$ with kernel of dimension 1, say that $I$ is a maximal indexing set of linearly independent columns, the indices of some set of columsn that give a basis for the image. Near $A$, $I$ still works as a basis for the image, because of continuity of the determinant. The Plucker coordinates for the image are the determinants of the $(n-1) \times (n-1)$ minors of $A_I$ (columns of $A$ corresponding to indices $I$), so they are changing algebraically.
This means you could in principle use the determinant-derivative formula to control how fast the kernel is moving along a path of matrices $A$ (with all one dimensional kernels).
I'm not sure that this could ever be useful in an application where you couldn't already compute the kernel of $A$ (i.e. the stationary distribution of $\lambda P + (1 - \lambda) Q = I + A$).
For example, plugging the case $A_m(t) = [tP + (1 - t) Q]_m$ (some corresponding minor of the mixture) into the determinant derivative formula, $d/dt (det(A(t)) = det(A(t)) tr( A^{-1}(t) d/dt A(t))$, we get $d/dt ( det(A(t)) = det ( A(t) ) tr( (A(t))^{-1} [P - Q]_m)$. To see how the speed at which $ker( tP + (1 - t)A)$ is moving around around on the Grassmannian, you have to compute a vector $(A_m(t))_{m \in minors(I)}$, for some local choice of $I$, normalize to make this to make it a path on the sphere, and then compute the derivative of the resulting path.
This seems kind of useless, but I guess I'll leave it up.
