TITLE: Shannon entropy and doubly stochastic matrices
QUESTION [6 upvotes]: Suppose that $A$ is a stochastic matrix. We know that if $A$ is doubly stochastic, then $H(Ap)\geq H(p)$ where $H$ is Shannon entropy and $p$ is a probability vector. Is the converse true? i.e., if $H(Ap)\geq H(p)$ then $A$ is doubly stochastic.

REPLY [4 votes]: Consider any $n \times n$ left stochastic matrix $A$, i.e. each column sums up to $1$. We argue that if $H(Ap) \geq H(p)$ for all probability distributions $p$, then $A$ is doubly stochastic.
Take $p$ to be the uniform distribution. Then $H(Ap) \geq H(p)$ implies that $Ap = p$, since the uniform distribution on $[n]$ is the unique maximizer of Shannon entropy among all probability distributions on $[n]$.
Since $p = \frac{1}{n} 1$ (the all ones vector), we have that $A1 = 1$ -- all rows sum up to $1$. Thus, $A$ is doubly stochastic.