the vector in $\R^n$ with all entries equal to $1$. The double stochasticity of $A$ implies $$A {\bf 1}= {\bf 1}, \qquad {\bf 1}^T A = {\bf 1}^T.$$ Note that multiplication by a doubly stochastic matrix $A$ preserves the average of the entries of a vector, i.e., for any $x\in\rn$, there holds $$\overline{Ax} = \frac{1}{n}\, {\bf 1}^T Ax
=\frac{1}{n}\, {\bf 1}^T x = \bar{x}.$$ We now write the quadratic form $V(x)-V(Ax)$ explicitly, as follows: $$\begin{aligned}
 V(x) - V(Ax) & = &  (x - \bar{x} {\bf 1})^T (x-\bar{x}{\bf 1}) -
(Ax-\overline{Ax}{\bf 1})^T (Ax - \overline{Ax}{\bf 1}) \nonumber \\
& = &  (x - \bar{x} {\bf 1})^T (x-\bar{x}{\bf 1}) - (Ax-\bar{x}A{\bf
1})^T (Ax - \bar{x} A {\bf 1}) \nonumber \\ & = &  (x - \bar{x} {\bf
1})^T (I-A^T A) (x-\bar{x}{\bf 1}). \label{vdec} \end{aligned}$$

Let $w_{ij}$ be the $(i,j)$-th entry of $A^T A$. Note that $A^T A$ is symmetric and stochastic, so that $w_{ij}=w_{ji}$ and $w_{ii} = 1 - \sum_{j \neq i} w_{ij}$. Then, it can be verified that $$A^T A  = I - \sum_{i<j} w_{ij} (e_i - e_j) (e_i -
e_j)^T, \label{asquareform}$$ where $e_i$ is a unit vector with the $i$-th entry equal to 1, and all other entries equal to 