Let $x, y$ be vectors in $\mathbb{R}^n$. Can it be shown that if the inner product satisfies $\langle x, y\rangle > 0$, then $\langle Ax, y\rangle > 0$ whenever $A$ is positive-definite?
[Math] Weighted inner product
inequalityinner-productslinear algebra
Best Answer
No, this won't work. Take $x = \binom 2 {-1}$ and $y = \binom 1 {1}$ in $\mathbb R^2$ and $$A = \left(\begin{matrix} 1 & 0 \\ 0 & 10 \end{matrix} \right).$$ Certainly $A$ is positive definite and $\langle x , y \rangle = 1 > 0$ but $\langle Ax, y \rangle = -8 < 0.$ Higher dimensional counterexamples are easy to construct as well.