TITLE: $L^2$ Bounds for Markov Chains.
QUESTION [9 upvotes]: Consider a non-negative, square stochastic $n \times n$ matrix $P$ (rows sum to one, $P$ is ergodic).
We are interested in characterizing the set of $n \times n$ invertible matrices $A$ such that we have:
$\| APA^{-1} \|_2 \leq 1$
Here $ \| \cdot  \|_2$ means a matrix norm induced by the vector norm $L_2$.
One example is $\operatorname{diag}(\xi)^{1/2}$, where $\xi$ is the vector representing the stationary distribution and $\operatorname{diag}$ is an operator constructing a diagonal matrix with zeros off the diagonal.
I am interested in a rule defining all (or at least a large class) of valid choices for $A$.

REPLY [2 votes]: @ ziutek , a stochastic matrix does not necessarily admit a stationary distribution. cf. below.
Note that $||APA^{-1}||_2=\sqrt{\rho (MM^T)}=\max_i \sigma_i$ where $M=APA^{-1}$ and $(\sigma_i)_i$ are the singular values of $M$ in decreasing order. Moreover $\rho(M)=\rho(P)=1$ and (it is true for every matrix) $\sigma_1\geq \rho(M)$. Thus your question is "find $A$ s.t. $||M||_2=\rho(M)$".
In this file $\| AB\|_\square \leq 1$ implies $\| BA\|_\triangle \leq 1$ , the following result is shown:
Prop. (user1551): there is a sub-mult. norm $||.||$ s.t. $ρ(U)=||U||$ iff the eigenvalues of $U$ of maximal modulus are semi-simple. 
Necessarily, the eigenvalues of $P$ of  modulus $1$ must be semi-simple. That is the case when $P$ is irreducible (because they are simple) that we'll assume in the sequel (that does not imply that a stationary distribution exists !). We may assume that $P=diag(T,N)$ where $\rho(T)<1$ and $N^h=I$ where $h$ is the period of $P$. Take $A=diag(B,C)$ with $B$ as in the proof above cited; then $||APA^{-1}||_2=||CNC^{-1}||_2$ and the condition is $||CNC^{-1}||_2=1$. I think that we MUST choose $C$ as follows: $C=RS$, a REAL matrix, s.t. $S$ (a complex matrix) diagonalizes $N$ and $R$ is unitary. 
EDIT: A mistake. What I wanted to say is that a stationary distribution is not necessarily a limit distribution.