Minimum of functional $x\mapsto\langle Ax,x\rangle -\langle b,x\rangle$ when $A$ is not symmetric

Question

In optimisation, a classic problem is the following:

Given positive definite matrix $A$ and a vector $b \in\Bbb{R}^n$, find the minimum of the functional $$x\mapsto\langle Ax,x\rangle -\langle b,x\rangle$$ in order to find a solution for the equation $Ax=b$.

I was wondering how this is useful in real life, since being positive definite seems to be a pretty strong condition. There must be some way this is generalized to the case where $A$ is just invertible, but I don't see how.

Answer 1

If $A$ is invertible but not positive definite, it has an eigenvalue $\lambda < 0$. If $v$ is the corresponding eigenvector, then $\lim_{c \to \infty} (cv)^T A (cv) + b^T(cv) = \lim_{c \to \infty} \lambda c^2 + c (b^Tv) = -\infty$.

If $A$ is not symmetric, you can replace it with $0.5(A+A^T)$ and get the same function values (and same optimum).