$\mathcal{l}_1$ norm replaced with reweighted $\mathcal{l}_2$ In this paper, (section 2, page 2)$\mathcal{l}_1$ norm is replaced with reweighted $\mathcal{l}_2$ in an optimization problem. I don't understand how $\lVert x\rVert_1$ is replaced with $x^TWx$ and ow the solution has changed to its weighted version. 
$$\min_x \lVert x\rVert_1 \quad \text{subject to} \quad Ax=b $$
$$\Downarrow$$
$$x^TWx \quad \text{subject to} \quad Ax=b$$
$$\Downarrow$$
$$x^{k+1}=(W^k)^{-1}A^T(A(W^k)^{-1}A^T)^{-1}b$$
 A: The formula for $x^{k+1}$ comes from constraint optimization, in particular, Lagrange multipliers: given a problem of the form
$$
\min f(x)\\
s.t. g(x)=0
$$
introduce the Lagrangian $\mathcal{L}(x,\lambda)=f(x)+\lambda g(x)$. Then, a solution to the minimization problem is a stationary point for the Lagrangian, that is,
$$
\partial_x \mathcal{L}(x,\lambda)=\partial_x f(x)+\lambda \partial_x g(x) =0\\
\partial_\lambda \mathcal{L}(x,\lambda)= g(x)=0.
$$
In your case, the Lagrangian is $\mathcal{L}(x,\lambda)=\frac{1}{2}x^TWx+\lambda^T(Ax-b)$ (the factor 1/2 does not change the point where the minimum is achieved, but simplifies calculations later), which leads to the system
$$
W^Tx+A^T\lambda=0\\
Ax=b,
$$
Now, the first equation can be solved for $x$, giving $x=-W^{-1}A^T\lambda$, which plugged in the second gives $-AW^{-1}A^T\lambda=b$, or $\lambda=-(AW^{-1}A^T)^{-1}b$. Plug this back in the first equation and you get
$$
W^Tx-A^T(AW^{-1}A^T)^{-1}b=0\Rightarrow x=(W^T)^{-1}A^T(AW^{-1}A^T)^{-1}b
$$
which is the formula you see.
