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 $K.^2$?

## 1 Answer

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Theorem 3 in "On majorization and Schur products" by Bapat and Sunder provides a stronger result. Take $A=B=K$ and specialize to $k=1$ in the majorization.