# How to solve ADMM for TV Minimization Problem For Different Sizes $A$ and $x$ in $Ax=b$

I have matrix $$A$$ that is $$(M \times M)$$ square matrix, $$x$$ $$(M \times N)$$ matrix, $$b$$ is $$(M \times N)$$ matrix. Knowing $$A$$ and $$b$$ I would like to get the $$x$$ from the equation $$Ax=b$$. $$N=p \times q$$, so consider $$x$$ as an $$M$$ number of $$p \times q$$ pixel images. These images are sparse when we take total variational $$TV(x)$$. I would like to solve the following minimization problem for $$x$$;

Define an operator $$g(x)=\parallel x^T\parallel_2,$$ or $$g(x)=\parallel x^T\parallel_1,$$ that maps $$M \times N \mapsto N \times 1$$.

$$\min \frac{1}{2}\parallel Ax-b\parallel^2_2+k\parallel z\parallel_1,$$ subject to $$Fg(x)-z=0$$.

$$F$$ is the difference matrix to take numerical gradient of pixels.

Unlike in the deblurring+denoising problem, my matrix $$A$$ and $$x$$ are in different sizes.

The ADMM (Alternating Direction Method of Multipliers) solution to my minimization problem is given as:

$$x^{k+1}=(A^TA+pF^TF)^{-1}(A^Tb+pF^T(z^k-u^k))$$

$$z^{k+1}=S_{t/p}(Fx^{k+1}+u^{k})$$

$$u^{k+1}=u^k+Fx^{k+1}-z^{k+1}$$

When calculating $$x^{k+1}$$ the matrix dimensions are not compatible in my case. Gradient matris should be applied to total number of pixels $$N$$, $$F$$ matrix should have $$N$$ columns. So the term $$(A^TA+pF^TF)$$ is not proper way to add 2 matrices.

How can I overcome this and find a solution to my problem?

• Which matrix multiplication exactly causes the dimension mismatch? At first glance they do look okay. $p$ is a scalar, right? Mar 3, 2020 at 7:51
• Yes $p$ is scalar. The dimension mismatch is addition of $A^TA$ and $F^TF$. $F$ is a scalar gradient operator which should be applied to all pixels of $x$. Since total number of pixels in $x$ is $N$, $F$ should be $NxN$. Mar 3, 2020 at 11:10
• @Florian what do you think about my question? Mar 4, 2020 at 19:24
• I don't understand the source of mismatch. $A$ multiplies $x$ in your cost function and $F$ multiplies $x$ in your constraint. Hence $A$ and $F$ have the same number of columns. Hence $A^T A$ and $F^T F$ are exactly the same size. Mar 5, 2020 at 8:11
• A is not applied to pixels in x. Thats why it is $M \times M$ @Florian Mar 5, 2020 at 13:44

### The Error in the Model

The problem is in the dimensions of the Linear Operator $$A$$ in your model compared to the Data Matrix $$X$$. The number of columns of the matrix $$A$$ must match the number of pixels in each column of $$X$$ (Each image). While in your case it matches the number of images.

### Matrix Form of 2D Linear Operator

Let's try to build the model correctly. Assume our data (Images) is given by the set of 2D matrices $${\left\{ {X}_{i} \right\}}_{i = 1}^{m}$$ where $${X}_{i} \in \mathbb{R}^{p \times q}$$. Given a Linear Operator $$\mathcal{A}: \mathbb{R}^{p \times q} \to \mathbb{R}^{k \times l}$$ and we have $${B}_{i} = \mathcal{A} \left( {X}_{i} \right)$$.

In order to create the matrix form we should represent the Linear Operator by the matrix $$A \in \mathbb{R}^{\left( k l \right) \times \left( p q \right)}$$ which is the matrix which applies the operator on the Column Stacked (See Vectorization Operator) vectors. So we have $$\boldsymbol{x}_{i} = \operatorname{Vec} \left( {X}_{i} \right)$$ and $$\boldsymbol{b}_{i} = \operatorname{Vec} \left( {B}_{i} \right)$$.

By defining $$X = \left[ \boldsymbol{x}_{1}, \boldsymbol{x}_{2}, \ldots \boldsymbol{x}_{m} \right]$$ and $$B = \left[ \boldsymbol{b}_{1}, \boldsymbol{b}_{2}, \ldots \boldsymbol{b}_{m} \right]$$ we have $$A X = B$$.

With this definition everything will work as expected. Same logic will work for $$F$$.

Remark

If you want $$F$$ to work along the rows of $$X$$ then set $$Z = F {X}^{T}$$.

### ADMM for the Vector Case

First, pay attention that the ADMM works on vectors (You can work with Matrices but then you need to update the Prox operations accordingly.

So $$\boldsymbol{x} \in \mathbb{R}^{n}, A \in \mathbb{R}^{m \times n}, \boldsymbol{b} \in \mathbb{R}^{m}$$ and $$F \in \mathbb{R}^{o \times n}$$ where $$o$$ is set to match the operation of TV on the image columns and rows which $$\boldsymbol{x}$$ is a column stack of.

So now the terms do align propely: $${\left( {A}^{T} A + p {F}^{T} F \right)}^{-1} \in \mathbb{R}^{n \times n}$$ and $$\left( {A}^{T} \boldsymbol{b} + p {F}^{T} \left( \boldsymbol{x} - \boldsymbol{u} \right) \right) \in \mathbb{R}^{n}$$. So the calculation of $$\boldsymbol{x}^{k}$$ is well defined.

Given that $$\boldsymbol{z} \in \mathbb{R}^{o}$$ and $$\boldsymbol{u} \in \mathbb{R}^{o}$$ their calculation is also well defined.

I guess something in your code doesn't match. But if the code have the terms $$A \boldsymbol{x} - \boldsymbol{b}$$ and $$F \boldsymbol{x}$$ well defined then everything else will work.

• Yes with dimensions you gave it should work. But I have different dimensions. So how can I solve the problem with matrices that i stated in my question? Mar 3, 2020 at 10:48
• It works with your dimensions the same way. Again, once you have $A X - B$ working and $F X$ well defined all the rest will work.
– Royi
Mar 3, 2020 at 10:51
• Total number of pixels in $x$ is $N$ and I need to apply F to take gradient of pixels values in $x$. So $F$ should be $NxN$ Matrix which cannot add to $A^TA$ because it is $MxM$ Mar 3, 2020 at 10:58
• This is what you miss. The right way to do this is column stack the image. If the image is $p \times q$ then your vector $\boldsymbol{x}$ has the length $n = p \times q$. From there follow what's in my answer.
– Royi
Mar 3, 2020 at 11:27
• If the image is $p \times q$ my $x$ is $M \times p \times q$. Then taking $N=p \times q$ I can write $x$ as $M x N$. Am I missing anything? Mar 3, 2020 at 11:37