Eigengap of $SS^T$ vs. $\frac{1}{2}(SS + S^TS^T)$, where S is doubly stochastic & irreducible Suppose $S$ is doubly stochastic irreducible matrix, and consider the two matrices $G=SS^T$ and $A=\frac{1}{2}(SS+S^TS^T)$. Note that in some sense $G$ is the "geometric mean" of $SS$ and $S^TS^T$, while $A$ is the arithmetic mean.
Both $G$ and $A$ are symmetric, doubly stochastic and irreducible, and thus have dominant eigenvalue $1$ relative to the eigenvector $1$ (the eigenvector with all components equal to $1$). I am interested in the second largest eigenvalue, and in particular in understanding whether I can say anything about whether it's larger in $G$ or in $A$. 
Note that applying Rayleigh's principle, we can write the second eigenvalue of a symmetric irreducible double stochastic matrix $M$ as $\sup_{x:~1^Tx=0} \frac{x^TMx}{x^Tx}$ (we are taking the sup over all vectors x that are orthogonal to the dominant eigenvector). Perhaps some generalization of the geometric-arithmetic mean inequality can be brought into play to prove that the second largest eigenvalue of $A$ is never smaller than that of $G$?
 A: I'll prove that the second largest eigenvalue of $G$ is not less than that of $A$. This is quite easy and almost proved in the comments.
Since $x^T(S^T - S)^T(S^T - S)x = ||(S^T-S)x||^2$ for all $x$, we have
$$
x^T(S^T S + SS^T)x \ge x^T(S S + S^T S^T)x = 2x^T A x.
$$
Therefore, 
$$
\sup_{x: 1^T x = 0, ||x||=1}  x^T(S^T S + SS^T)x \ge 2\sup_{x: 1^T x = 0, ||x||=1} x^TAx.
$$
On the other hand,
$$
\sup_{x: 1^T x = 0, ||x||=1}  x^T(S^T S + SS^T)x\le \sup_{x: 1^T x = 0, ||x||=1}  x^TS^T Sx + \sup_{x: 1^T x = 0, ||x||=1}  x^TS S^Tx.
$$
The right-hand side is the sum of second largest eigenvalues of $SS^T$ and $S^TS$. But the latter are equal, which implies the claim.
Similarly it can be shown that the sum of first $k$ largest eigenvalues of $G$ is not less than that of $A$ for any $k\ge 1$ and any, not necessarily doubly stochastic, matrix $S$. The proof is based on this generalization of Rayleigh principle: for a symmetric matrix $M$ the sum of $k$ largest eigenvalues is
$$
\sup_{\text{ONS }x_1,x_2,\dots,x_k} \sum_{i=1}^k x_i^T M x_i,
$$
where ONS stands for "orthonormal system".
