Let $A$ be an invertible $n \times n$ matrix and $v \in \mathbb R^n$. We look for a minimizer $x^*$ of $F(x) = x^\top A x - 2v^\top x$. It is known that $x^* = A^{-1}v.$ Until now, everything is clear to me. Let us now suppose that the symmetric part of $A$, i.e., $A_S=(A^\top+A)/2$, is invertible, too. It is known that $x^\top A x = x^\top A_S x$, since the skew-symmetric part gives zero. Therefore, $$ F(x) = x^\top A x - 2v^\top x=x^\top A_S x - 2v^\top x, $$ and minimizing the second or the first quadratic forms should be the same. Therefore, we should have $x^*=A_S^{-1}v$ and finally $$ (A^{-1} -A_S^{-1})v =0, $$ which implies that either $A^{-1} -A_S^{-1}=0$ or that $v \in \mathrm{Ker}(A^{-1}-A_S^{-1})$.
I feel like I'm getting something wrong here (first possibility is for sure not true if $A$ is not symmetric as $A_S^{-1}$ will be symmetric, second possibility sounds odd since $v$ can be any vector)