Given two distinct vectors $u$ and $v$ in $\mathbb{R}^n$ such that $u \neq \lambda v$ for any $\lambda \in \mathbb{R}$, find a symmetric positive definite matrix $A \in \mathbb{R}^{n\times n}$ such that $u^TAv=0$. The vectors $u$ and $v$ are thus $A$-orthogonal or $A$-conjugate vectors. Is this even possible ? I am currently not worried about uniqueness of $A$, although that is also a good question to ponder.
Thanks to user little0 for pointing out $u = \lambda v$ situation.
If $u = v$ (and both are nonzero), then $u^T A v = u^T A u > 0$ for all symmetric positive definite matrices $A$.
So, in that case it is not possible to find a symmetric positive definite $A$ such that $u^T A v = 0$.
The same conclusion holds if $u = \lambda v$ for some scalar $\lambda$, and both $u$ and $v$ are nonzero. In this case, if $A$ is symmetric positive definite, then $$ u^T A v = \lambda v^T A v $$ which is nonzero because $v^T A v > 0$ and $\lambda \neq 0$.