Can a stochastic matrix be written as $V^{-1} D V $? V is an invertible matrix and D is diagonal. I think so but I can't think of a good proof.
Also, the left eigenvectors and right eigenvectors are not necessarily the same so that the right eigenvectors are only orthogonal with respect to a measure/matrix. That is, if $\{v_i\}$ are right eigenvectors then $v_i^\dagger V^\dagger V v_j=\delta_{ij}$. Is there an easy way to find out what this matrix is?
For clarity, a stochastic matrix, $M$, is one for which $\sum_i M_{ij}=1$ for all $j$.
I want to partially answer my own questions. If the stochastic matrix satisfies detailed balance then the matrix that we take the inner product respect to is given by $\sum_{ij}\delta_{ij}\frac{1}{\pi_i}$. where $\pi_i$ is the equilibrium distribution. Detailed balance says that $M_{ij} \pi_j=M_{ji}\pi_i$. This is because the matrix $\sum_{ij}\delta_{ij}\frac{1}{\pi_i} M$ is Hermitian and thus has real eigenvalues. Hence $v_i^\dagger \sum_{ij}\delta_{ij}\frac{1}{\pi_i} M v_j=\lambda_i v_i^\dagger \sum_{ij}\delta_{ij}\frac{1}{\pi_i} v_j=\lambda_j v_i^\dagger \sum_{ij}\delta_{ij}\frac{1}{\pi_i} v_j$ and so if the eigenvalues differ, then the eigenvectors are orthogonal.