Is there regularization term which is simply a linear function of $\beta$? there have been a lot of discussions on regularization in the linear regression context:
$\min( (A*\beta -y)^2 + \lambda |\beta|_{0,1,2})$
L0 regularization is counting the number of non-zero beta
L1 is absolute value
L2 is square
I am curious why I have never heard about regularization term that is simply a linear function of $\beta$
$\min( (A*\beta -y)^2 + \lambda \beta)$
My questions are:
1) most importantly, is there any theoretical flaw of the linear regularizer?
2) if not, what's the best way to solve this?    
 A: $\beta$ is a vector, and we need to map it to a scalar to minimize the weighted sum of the error (or negative log-likelihood) and this scalar.
The canonical way to go about this is to use a norm of $\beta$. As you write, different norms imply that the minimizer has different properties.
A: If you meant to use $\lambda \sum \beta$ as your penalty term, this will have the problem that your algorithm will try to make the $\beta$ values as negative as possible consistent with a reasonable fit to the data (not as small in magnitude as possible, which is probably what you want). As @StephanKolassa points out in a comment above, letting $\beta \to -\infty$ would likely minimize the criterion. (A tiny bit of analysis would be required to actually prove this; you'd need to show that (for a particular value of $\lambda$) the magnitude of the derivative of $\|X\beta-y\|^2$ is smaller than the derivative of $\lambda \sum \beta$ ...)
A: Stephan Kolassa has a nice answer.
In addition, there are some problems in your math. 
The first problem is mixing $A$ and $\beta$. 
A better notation can be
$$
\text{minimize}~~ \|X\beta-y\|^2+\lambda\|\beta\|^2
$$
or 
$$
\text{minimize}~~ \|Ax-b\|^2+\lambda\|x\|^2
$$
If data matrix is represented with $A$ then $x$ is paired with. And if data matrix is represented with $X$, $\beta$ is paired with. In both settings, $\|\cdot\|$ represent vector norm.
In addition, people usually do not use $*$ on matrix multiplication. 
Finally, in your equation, It is hard to see $(A*\beta -y)^2 $ is a scalar. If it is a vector $\min$ on a vector is not clearly defined. I think this is the root of your confusion. As Stephan Kolassa mentioned, usually, objective function returns a scalar.
