Direction of steepest descent and minimization? I have the following linear function: $min$ 1/2 $<x, x>$ + $r^Tx$ for every x belonging to $R^n$, $r^Tx$ belongs to $R^n$ Now, = $x^TAy$ and A is symmetric positive definite. = $x^TAy$ is the scalar product in $R^n$ and the dimensions of A are n$x$n. The task is to derive a linear function for x, i.e. solve the linear optimization problem. From what I understand, I need to solve the above-mentioned function for , i.e. for $x^TAy$. Solving it for $x^TAy$ is just a standard quadratic minimization problem, but I am concerned for the y, i.e. a different variable. And how can I show that x is perpendicular to the kernel of $r^T$. It is given that ker($r^T)x$ belongs to $R^n$ such that $r^Tx = 0$?

The problem is $$ \min_{x\in\mathbb{R}^n} f(x) = \min_{x\in\mathbb{R}^n} \frac{1}{2} x^T A x + r^T x,$$ which has gradient $$ \
abla f(x) = A x + r.$$ The necessary condition $\
abla f(x_*)=0$ gives us that $Ax_*=-r$, which has a unique solution. Since the Hessian $A$ is positive definite we are guaranteed the existence and uniqueness of the solution to $A x_* = -r$, and that the point $x_*$ is a minimizer.

In the latter part of your question, do you mean perpendicular in $\mathbb{R}^n$ or perpendicular w.r.t. your inner product, i.e., $=x_*^T A \ker(r) = 0$? Perhaps knowing $x_* = -A^{-1} r$ is enough to let you finish the problem.