In solving the problem:
$$\underset{x}{\text{minimize}} \quad ||Ax - b||_{1}$$
We can formulate it as an LP by noting that:
$$||Ax-b||_{1} = \sum_{i} |a_{i}^{T}x - b_{i}|$$
In Boyd's book, they rewrite it as:
$$\underset{x}{\text{minimize}} \quad 1^{T} \mathbf{\delta}$$ $$\text{subject to} \quad -1^{T} \mathbf{\delta} \leq Ax-b \leq 1^{T}\mathbf{\delta}$$
But shouldn't $\mathbf{\delta} \geq 0$? In other words, shouldn't the LP be:
$$\underset{x}{\text{minimize}} \quad 1^{T} \mathbf{\delta}$$ $$\text{subject to} \quad -1^{T} \mathbf{\delta} \leq Ax-b \leq 1^{T}\mathbf{\delta}\quad \mathbf{\delta} \geq 0$$
Or is that distinction somehow implicit? Or does it even matter?
You mention that Boyd's book states: $$ -1^T\delta \le Ax-b \le 1^T\delta$$
I think that should be: $$ -\delta \le Ax-b \le \delta$$
From this "sandwich equation" we can see that we require $-\delta \le \delta$ which implies $\delta \ge 0$.
So, $\delta \ge 0$ holds automatically.