If $A$ is symmetric p.s.d, and $\langle u,v\rangle \leq 0$, is $u^{T} A v \leq 0$?

Question

Sorry for the lack of work, I'm not sure how to go about proving or disproving it.

Answer 1

This is not true. Here is a counterexample for $\mathbb{R}^2$.

Consider $A=\left(\begin{matrix} 0.5 &1\\ 1 & 2 \end{matrix}\right)$.

Then $A$ is positive semidefinite since its eigenvalues are $0$ and $5/2$.

Let $u=(1,0)^T$ and $v=(-1,1)^T$. Then $\langle u,v\rangle = -1 \leq 0$ but $u^T A v=0.5 \geq 0$