Exponential convergence of a Markov chain We are learning about the exponential  convergence of a Markov chain.
We learned that if a Markov chain(represented by the transition matrix $P$) is a-periodic, irreducible and reversible then it converges exponentially and the rate of convergence(in term of the total variation norm)is the second largest eigenvalue of $P$.
I wonder if this is also true if $P$ is just a-periodic, irreducible and diagonalizable. If $P$ is diagonalizable then I can write $ P = S D S^{-1}  \implies P^n  = S D^n S^{-1}   $ where $D$ is a diagonal matrix of the eigenvalues of $P$.
We know that the magnitude of all eigenvalues is less than 1 except a single eigenvalue with a value of 1 which means that $ P^n  = S D^n S^{-1}   $ converges in the rate of the second largest eigenvalue of $P$.
Hope that someone can clear my confusion.
 A: Suggestion for this kind of problems: try to visualize the action of the linear operator in its eigenbasis.

Say $P$ is a diagonalizable $m\times m$ irreducible stochastic matrix.
Let $\{\mathbf p,\mathbf v_2,\cdots\mathbf v_m\}$ be the $m$ eigenvectors associated with the eigenvalues $\{1,\lambda_2,\cdots\lambda_m\}$. As you state in the question, $|\lambda_j|<1 \quad \forall j$. Degeneracy in $\lambda_j$ allowed.
Apply $P^n$ to some probability distribution $\mathbf q = c_1\mathbf p+c_2\mathbf v_2 + \cdots + c_m\mathbf v_m$.
$$P^n \mathbf q = c_1\mathbf p+c_2 \lambda_2^n\mathbf v_2 + \cdots + c_m\lambda_m^n\mathbf v_m$$
As you can see, the distribution is converging exponentially to the steady-state distribution $\mathbf p$, with convergence rate determined by the largest $\lambda_j$ (second largest eigenvalue).

In a left stochastic matrix each column sums up to $1$, $\mathbf 1^T P=\mathbf 1^T$.
Thus, $P \mathbf v_j = \lambda_j \mathbf v_j \Leftrightarrow \mathbf 1^T \mathbf v_j = \lambda_j \mathbf 1^T \mathbf v_j \Leftrightarrow \mathbf 1^T \mathbf v_j =0$ since $\lambda_j \neq 1$.
Together with, $\mathbf 1^T \mathbf q =1$ and $\mathbf 1^T \mathbf p =1$ we find $c_1 =1$.
