Let $A$ be an $n\times n$ symmetric matrix, let $b$ be an $n$-vector, let $c \in \mathbb{R}$ and set $Q(x) = 1/2 x^T Ax-x^T b+c$. Prove that $x_0$, defined as a solution to $Ax_0=b$ is a critical point.
Additionally I am wondering what are the conditions on A such that $x_0$ is the minimizing point? What happens in other cases?
The critical points are defined as the points where the derivative becomes zero. In this case, we calculate $$ DQ = x^TA - b^T $$ Alternatively, you can "complete the square" to find that $$ Q(x) = \frac 12 (x - x_0)^TA(x - x_0) + c' $$ for some vector $c'$.
We find that $x_0$ is a minimizing point when $A$ is positive definite, $x_0$ is a maximizing point when $A$ is negative definite. Whenever $A$ is non-singular but falls into neither of the previous cases, $A$ will be a proper saddle point. When $A$ is singular, $Ax_0 = b$ no longer has a unique solution (if such a solution exists).