Consider the problem $$\min f(x)\\Ax=b$$ where $f:\mathbb{R}^n\to\mathbb{R}, f\in C^1, A\in\mathbb{R}^{m\times n} b\in \mathbb{R}^m$, $m<n$ and $rank(A) = m$. Let $\overline{p}$ be the solution of $$\min ||\nabla f(x) -p||\\Ap=0$$ Find $\overline{p}$ and interpret geometrically
I'm learning to solve $\min f(x)$ subject to $Ax = b$. In theory we'd simply transform this to an unconstrained problem by analyzing $\phi(\gamma) = f(\overline{x}+Z\gamma)$ where $\overline{x}$ is a solution for $Ax=b$, $Z$ is the matrix of the basis of $\ker(A)$ and $\gamma$ is our variable of unscontrained optimization. We must now find $\min \phi(\gamma)$ subject to the entire $\mathbb{R}^{\mbox{something}}$. For this, we calculate $\gamma$ that solves $\nabla \phi(\gamma) = Z^T\nabla f(\overline{x}+Z\gamma)=0$ and see if it solves $\nabla^2 \phi(\gamma) = Z^T\nabla^2 f(\overline{x}+Z\gamma)Z>0$.
So let's apply this reasoning to our second problem (what is the first one for anyways?):
Call $h(p) = ||\nabla f(x) -p||$ and thus
$$h(p) = ||\nabla f(x) -p|| = \sqrt{(\frac{\partial f}{\partial x_1}-p_1)^2 + \cdots +(\frac{\partial f}{\partial x_n}-p_n)}\implies \\ \frac{\partial h}{\partial p_i} = \frac{2(\frac{\partial f}{\partial x_i}-p_i)(-1)}{2\sqrt{(\frac{\partial f}{\partial x_1}-p_1)^2 + \cdots +(\frac{\partial f}{\partial x_n}-p_n)}}$$
so
$$\nabla h(p) = \frac{p-\nabla f(x)}{||\nabla f(x) -p||}$$
Our $\overline{x}=0$ and thus we have
$$\phi(\gamma) = h(\overline{x}+Z\gamma) = h(Z\gamma) = \frac{Z\gamma-\nabla f(x)}{||\nabla f(x) -Z\gamma||}$$
we must find $\gamma^*$ such that $\frac{Z\gamma^*-\nabla f(x)}{||\nabla f(x) -Z\gamma^*||}=0$ and then test the point $x^* = \overline{x} + Z\gamma^*$ in $\nabla^2 \phi(\gamma)$. How should I find the point $\gamma$ and how to find the hessian? It's going to be a big thing. Also what is the first problem doing there?