How to prove the following interesting fact about positive definite matrix?

Question

if $M$ is a real positive definite matrix, $S$ is a real vector, we define $\mathbb{sign}(S) = [\mathbb{sign}(s_1),...,\mathbb{sign}(s_n)]^T$. How to prove the fact that $S^TM\mathbb{sign}(S)$>0?

Answer 1

This is not necessarily true. For instance, consider $$ M = \pmatrix{1&-2\\-2&10}, \qquad S = \pmatrix{9\\1} \quad t > 0 $$ We compute $$ S^TM\operatorname{sgn}(S) = -1 < 0 $$

Answer 2

This is not true. Counterexample: $$ M=\pmatrix{ 16&-14&-11\\ -14& 35& -5\\ -11& -5& 41},\ S^T=(5,1,1),\ S^TM\operatorname{sign}(S)=-4. $$ One can easily verify that $M$ is positive definite using Sylvester's criterion.

Remark. Let $S$ be entrywise positive. Then $S^TM\operatorname{sign}(S)=S^TM\mathbf1$. Since a positive definite matrix may contain negative off-diagonal entries, $M\mathbf1$ can have negative elements too. Therefore it is implausible that $S^TM\mathbf1$ is always nonnegative.