Quadratically constrained linear program (QCLP) over $x$ with the linear constraint $x = Az$ I have a problem that looks very much like a norm-constrained linear program, but with an extra constraint that is unusual for me. The problem is the following. Given a matrix $A$ and a vector $w$,
$$ \min_{x \in \mathbb{R}^n} w^\top x $$
subject to
$$ \|x\|_2 = 1,$$
$$x_i \ge 0 ~\forall~ i,$$
$$x = Az.$$
Note that $A$ and $w$ are fixed, and $z$ is a free variable. It almost looks like a standard QCLP, but with the extra requirement  that $x$ be in the column space of $A$. We cannot assume that $A$ is invertible, but $z$ can take on any value in $\mathbb{R}^n$.
Of course we can substitute $Az$ for $x$ to get an equivalent problem with fewer constraints,
$$\min_{z \in \mathbb{R}^n} w^\top A z$$
subject to
$$\|Az\|_2 = 1,$$
$$Az \ge 0$$
(where the last inequality is element-wise), but I still don't know of methods that are designed to solve this problem, or even better if there is a closed-form solution.
Are problems like this well-studied in some corner of constrained optimization? I have seen similar ones, but not this one exactly.
 A: The projection to the column space of $A$ is done by $A A^\dagger x$ (where $A^\dagger$ is the pseudoinverse).  If $x$ is in the column space, $A A^\dagger x=x$.  Hence, the above constraint can be written as: 
$$(A A^\dagger -I)x=0$$.
A: We have the following (non-convex) quadratically constrained linear program (QCLP) in $\mathrm x, \mathrm y \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \mathrm w^\top \mathrm x\\ \text{subject to} & \| \mathrm x \|_2^2 = 1\\ & \mathrm x - \mathrm A \mathrm y = 0_n\\ & \mathrm x \geq 0_n\end{array}$$
where $\mathrm A \in \mathbb R^{n \times n}$ and $\mathrm w \in \mathbb R^n$ are given.

The easy case
Suppose that $\rm w \leq 0_n$ is in the column space of $\rm A$. Using the Cauchy-Schwarz inequality,
$$\mathrm w^\top \mathrm x \geq -\| \mathrm w \|_2 \underbrace{\| \mathrm x \|_2}_{=1} = -\| \mathrm w \|_2$$
where the minimum is attained at
$$\mathrm x_{\min} := - \frac{1}{\| \mathrm w \|_2} \mathrm w \geq 0_n$$
Hence, the inequality constraint $\mathrm x \geq 0_n$ is satisfied. If $\rm A$ is invertible, then
$$\mathrm y_{\min} := \mathrm A^{-1} \mathrm x_{\min} = - \frac{1}{\| \mathrm w \|_2} \mathrm A^{-1} \mathrm w$$
If $\rm A$ is non-invertible, then $\mathrm y_{\min}$ will be in the solution set of the linear system $\mathrm A \mathrm y = \mathrm x_{\min}$, which is non-empty due to the fact that $\rm w$ is in the column space of $\rm A$.

The general case
Let $\mathrm c := \mathrm A^\top \mathrm w$. We rewrite the original QCLP as follows
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm y\\ \text{subject to} & \| \mathrm A \mathrm y \|_2^2 = 1\\ & \mathrm A \mathrm y \geq 0_n\end{array}$$
which is non-convex. Replacing the equality constraint $\| \mathrm A \mathrm y \|_2^2 = 1$ with the inequality constraint $\| \mathrm A \mathrm y \|_2^2 \leq 1$, we obtain a convex optimization problem in $\mathrm y \in \mathbb R^n$
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm y\\ \text{subject to} & \| \mathrm A \mathrm y \|_2^2 \leq 1\\ & \mathrm A \mathrm y \geq 0_n\end{array}$$
Using the Schur complement, the inequality constraint $\| \mathrm A \mathrm y \|_2^2 \leq 1$ can be written as the following linear matrix inequality (LMI)
$$\begin{bmatrix} \mathrm I_n & \mathrm A \mathrm y\\ \mathrm y^\top \mathrm A^\top & 1\end{bmatrix} \succeq \mathrm O_{n+1}$$
Thus, we have the following semidefinite program (SDP)
$$\begin{array}{ll} \text{minimize} & \mathrm c^\top \mathrm y\\ \text{subject to} & \begin{bmatrix} \mathrm I_n & \mathrm A \mathrm y & \mathrm O_n\\ \mathrm y^\top \mathrm A^\top & 1 & 0_n^\top\\ \mathrm O_n & 0_n & \mbox{diag} (\mathrm A \mathrm y) \end{bmatrix} \succeq \mathrm O_{2n+1}\end{array}$$
Let $\mathrm y_{\min}$ be the minimizer. If $\| \mathrm A \mathrm y_{\min} \|_2 = 1$, we have solved the original (non-convex) problem.
