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The primary optimziation problem is shown as $$\min_A \|X-DA\|^2_F+\beta\|A\|_1,$$ where $X\in\Bbb{R}^{m\times p}$,$D\in\Bbb{R}^{m\times n}$ and $A\in\Bbb{R}^{n\times p}$ are three matrices. Let $A=A^1-A^2$, $A^1\in\Bbb{R}^{n\times p}$ and $A^2\in\Bbb{R}^{n\times p}$ are defined as $$ A^1_{ij} = \begin{cases} a_{ij}, & \text{if $a_{ij}>0$} \\ 0, & \text{otherwise} \end{cases} $$ $$ A^2_{ij} = \begin{cases} |a_{ij}|, & \text{if $a_{ij}<0$} \\ 0, & \text{otherwise} \end{cases} $$ where $a_{ij}$ is the $i$th row and $j$th column of $A$.

A paper said that this problem can be transformed as $$\min_{A^1,A^2} \|X-D(A^1-A^2)\|^2_F+\beta e_n^T(A^1+A^2)e_p$$ $$s.t. A^1\geq 0, A^2\geq 0.$$ where $e_n\in\Bbb{R}^{n\times 1}$ and $e_p\in\Bbb{R}^{p \times 1}$ are two all-one vectors. $A^1\geq 0$ and $A^2\geq 0$ denote that all elements in $A^1$ and $A^2$ are nonnegative.

My question is how to prove the primary problem and the transformed problem are equivalent?

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    You have that $A = A^1-A^2$ and isn't $||A||_1=\sum_{i,j}|a_{ij}|$ by definition? – 2017-02-22
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    @LeonSot It depends. Another possibility for $\|A\|_1$ is to denote the operator norm induced by $\mathbf R^n$'s $\|\cdot\|_1$. But I guess, you are right here. – 2017-02-22
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    @LeonSot should't $\|A\|_1=\max_j \sum_i |a_{ij}|$ ? – 2017-02-22

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