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Suppose I have a positive definite symmetric matrix $P$ and a nonlinear function $G(x)$ (non symmetric) which depends on a vector $x$ (both $P$ and $G(x)$ are $6x6$). Suppose also that I can guarantee that, if $x \in S$, G(x) will be negative semidefinite for every $x\in S$ . Can I say that the product $PG(x)$ is negative semi-definite whenever $x \in S$?

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Certainly If you assume $P$ ortogonal and simmetric $(P^{-1}=P^t $and $P^t=P)$ because in this case P admits a base of ortogonal eighenvectors $B=\{y_1,....y_n\}$ for the space in which theirs eighevalues are all strictly positive and if you consider $x\in S$ and $y_i\in B$ then

$<PG(x) y_i, y_i>=<G(x) y_i, P^t y_i>= $ $<G(x) y_i, P y_i>=\lambda_i <G(x) y_i,y_i> \leq0 $

because $\lambda_i>0$ and $<G(x) y_i,y_i>\leq0$

($P$ is definite positive and $G(x)$ is semi-definite negative)

If you consider $y\neq 0$ you have that there exist $\alpha_1,...,\alpha_n\in\mathbb R$ such that $y=\sum_{k=1}^n\alpha_ky_k$ and

$<PG(x) y,y>=\sum_{k=1}^n\alpha_k^2 \lambda_k<G(x) y_k,y_k>$ $\leq 0$

So in this case $PG(x) $ is always semi-definite negative.

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