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I'm trying to find a way to change $\min \|Ax-b\|_1 $, where x is free, to standard form. I'm extremely rusty on linear algebra, so help would be much appreciated.

The textbook gives a method to standardize $\min \|Ax-b\|$ where x>0, where we use the fact $\|z\| = \sum_i=1^m |z|, \|z\|_\infty = \max_{i = 1, \ldots, m} |z_i|, \|z\|_2 = \sqrt {\sum_{i=1}^m z_i ^2}$.

And so we define w such that $w_i = |(Ax-b)_i|$, $w_i \geq (Ax-b)_i$, $w_i \geq -(Ax-b)_i$, $w \geq (Ax-b)$, $w \geq -(Ax-b), x>0$.

Then rewriting w to $\min e^t w$ where $e^t w = \sum_{i=1}^m w_i$, and $w \geq Ax-b, \quad w \geq -(Ax-b), x\geq 0 , w \geq 0$

Below is my attempt at the original problem:

Since x is free, we will split $x = x^{+} + x^{-}$, such that $x^{+}>0$ and $x^{-}>0$. Does this mean our final standard form will just look like:

$\min w$ subject to $\begin{cases} w \geq A(x^{+} + x^{-}) - b\\ w \geq - (A(x^+ - x^-) + b)\\x^+ \geq 0, x^- \geq 0, w \geq 0 \end{cases}$

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    This (minimizing the $1$-norm) is a problem in *linear programming*, not so much linear algebra. I'll retag, but yes, the square norm minimization amounts to solving a linear system. – 2017-02-13

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From Boyd & Vandenberghe:

enter image description here

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    Ah thank you. If my understanding is correct, we will now be subject to $-t \leq A(x^+ - x^-) - b \leq$ for $x^+, x^- \in \mathbb{R}^n$ and $t \in \mathbb{R}^m$? – 2017-02-13
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    If you introduce $\rm x^+, x^-$, shouldn't they be nonnegative? – 2017-02-13