Derivative of $l_1$ norm I want to compute the following derivative with respect to $n\times1$ vector $\mathbf x$. 
$$g = \left\lVert \mathbf x - A \mathbf x  \right\rVert_1 $$

My work:

$$g = \left\lVert \mathbf x - A \mathbf x  \right\rVert_1  = \sum_{i=1}^{n} \lvert x_i - (A\mathbf x)_i\rvert = \sum_{i=1}^{n} \lvert x_i - A_i \cdot \mathbf x \rvert = \sum_{i=1}^{n} \lvert x_i - \sum_{j=1}^n a_{ij} x_j\rvert$$ 
So the $k$th element of derivative is:

$$\frac{\partial g}{\partial x_k} = \frac{\partial }{\partial x_k}\sum_{i=1}^n \lvert x_i - \sum_{j=1}^n a_{ij} x_j\rvert $$
$$= \frac{\partial }{\partial x_k}\bigg(\lvert x_1 - \sum_{j=1}^n a_{1j} x_j\rvert +\cdots+ \lvert x_k - \sum_{j=1}^n a_{kj} x_j\rvert + \cdots\lvert x_n - \sum_{j=1}^n a_{nj} x_j\rvert  \bigg)$$
$$ =-a_{1k}sign(x_1 - \sum_{j=1}^n a_{1j} x_j)-\cdots+(1-a_{kk})sign(x_k - \sum_{j=1}^n a_{kj} x_j)-\cdots -a_{nk}sign(x_n - \sum_{j=1}^n a_{nj} x_j)$$

And my questions:


Is this derivation correct?
How I can represent the answer compactly?
Can you introduce me a source to master this material?
Apart from a sign error, your result looks correct. The term with $(1-a_{1k})$ should have a positive sign. Also note that $\text{sgn}(x)$ as the derivative of $|x|$ is of course only valid for $x\neq 0$. If you take this into account, you can write the derivative in vector/matrix notation if you define $\text{sgn}(\mathbf{a})$ to be a vector with elements $\text{sgn}(a_i)$:

$$\nabla g=(\mathbf{I}-\mathbf{A}^T)\text{sgn}(\mathbf{x}-\mathbf{Ax})$$

where $\mathbf{I}$ is the $n\times n$ identity matrix.