Schur product theorem

The entrywise product of two positive semidefinite matrices is positive semidefinite. Writing each matrix as a sum of rank-one Gram matrices reduces the claim to $(u{u}^T)\circ (v{v}^T)=(u\circ v)(u\circ v{)}^T$.

Let real power series $f(t)={\sum }_{k\ge 0}{f}_k{t}^k$ and $g(t)={\sum }_{k\ge 0}{g}_k{t}^k$ converge on $[-1,1]$, with $f_k\ge |g_k|$ for every index including zero. If $X⪰0$ has entries there, applying $f$ within two diagonal blocks and $g$ across them preserves positive semidefiniteness. Indeed the coefficient matrix $H_k$ is a bipartite block-constant positive semidefinite matrix. Each $H_k\circ X^{\circ k}$ is positive semidefinite by the Schur product theorem, and their sum converges to the transformed matrix. Closedness of the positive semidefinite cone completes the proof. Convergence at one gives $\sum f_k<\infty$, so coefficient domination ensures absolute convergence of both series throughout the interval. Without the zero-index condition, $f=-1$, $g=0$ and $X=0$ give a counterexample.