Let $K$ be a covariance matrix. It is positive semidefinite, its diagonal elements are all 1, and its off-diagonals are between -1 and 1. Let $K.^2$ be its element-wise power (Hadamard power). Can we show that maximum eigenvalue of $K$ are great or equal than the maximum eigenvalue of $K.^2$? 



Maximum eigenvalue of Hadamard power of a positive semidefinite matrix
Theorem 3 in ["On majorization and Schur products"](https://www.sciencedirect.com/science/article/pii/0024379585901478/pdf?md5=f8866f6d5d885524d837052e8c42719d&pid=1-s2.0-0024379585901478-main.pdf) by Bapat and Sunder provides a stronger result. Take $A=B=K$ and specialize to $k=1$ in the majorization.


