Alternative matrix representation for translation The ''usual" way to write translation for $\textbf{v}\in \mathbb{R}^2$ is with the following $3\times3$ matrix
$$  \left( {\begin{array}{ccc}
   1 & 0 & x_{0} \\
   0 & 1 & y_{0} \\
   0 & 0 & 1 \\
  \end{array} } \right) \cdot \left( {\begin{array}{c}
   x \\
   y \\
   1 \\
  \end{array} } \right)= \left( {\begin{array}{c} x+x_{0}\\y+y_{0}\\1\end{array}}\right)$$
But also it seems to be possible to write this transformation with the following $2\times2$  matrix:
$$  \left( {\begin{array}{cc}
   1 & \frac{x_{0}}{y} \\
   \frac{y_{0}}{x} & 1 \\
   \end{array} } \right) \cdot \left( {\begin{array}{c}
   x \\
   y \\
    \end{array} } \right)= \left( {\begin{array}{c} x+x_{0}\\y+y_{0}\end{array}}\right)$$
I didn't find a source using this as a version of translation, what are the problems associated with this representation?  
 A: The matrix representing a map should depend on just that, the map (here the translation by $(x_0,y_0)$). If cannot depend on the input to the map (here the point $(x,y)$), because that is not any fixed quantity when regarding the map as a whole. So using $x,y$ in the matrix (or any other representation) of a translation makes no sense. Think of what would be your matrix for translation by $(2,3)$. You cannot write down your matrix for that concrete case, because it involves $x$ and $y$.
This is similar to why one cannot consider a polynomial function $x\mapsto ax^2+bx+c=(ax+b)x+c$ of degree$~2$ to instead be a degree $1$ polynomial function with coefficients $ax+b$ and$~c$. Polynomial coefficients should by definition not contain the argument $x$ of the polynomial function, and similarly matrix coefficients should not contain (components of) the argument the matrix is (presumably) going to be applied to.
If this still fails to convince, consider the following. Composition of linear maps should correspond to matrix multiplication. Indeed $AB\cdot v=A\cdot(B\cdot v)$ for matrices $A,B$ and a vector $v$ always holds (provided the sizes are appropriate for the products to be defined). If we apply this to the properly defined matrices for translation we get
$$
 \begin{pmatrix}
   1 & 0 & a_0 \\
   0 & 1 & b_0 \\
   0 & 0 & 1 \\
  \end{pmatrix}  \cdot \begin{pmatrix}
   1 & 0 & a_1 \\
   0 & 1 & b_1 \\
   0 & 0 & 1 \\
  \end{pmatrix} =\begin{pmatrix}
   1 & 0 & a_0+a_1 \\
   0 & 1 & b_0+b_1 \\
   0 & 0 & 1 \\
  \end{pmatrix}
$$
and indeed a translation by $(a_1,b_1)$ composed with a translation by $(a_0,b_0)$ gives a translation by $(a_0+a_1,b_0+b_1)$. However trying to do the same for your matrices (ignoring what the symbols $x,y$ mean)
$$
  \begin{pmatrix}
   1 & \frac{a_0}{y} \\
   \frac{b_0}{x} & 1 \\
   \end{pmatrix} \cdot
  \begin{pmatrix}
   1 & \frac{a_1}{y} \\
   \frac{b_1}{x} & 1 \\
   \end{pmatrix} =
  \begin{pmatrix}
   1+\frac{a_0b_1}{xy} & \frac{a_0+a_1}{y} \\
   \frac{b_0+b_1}{x} & 1+\frac{b_0a_1}{xy} \\
   \end{pmatrix}
$$
then the result does not even have the form of they matrix you proposed for a translation. You can see that "applying" this matrix to some vector $(x,y)$ does not have the same effect of applying the translation by $(a_1,b_1)$ and then the translation by $(a_0,b_0)$ to the result. This is so in spite of the fact that $AB\cdot v=A\cdot(B\cdot v)$, because the way you applied the matrix does not just involve matrix$\times$vector multiplication, but also a substitution of values for $x,y$ into the matrix, which are not the same values when doing the second translation.
I should note that representing translations by matrices requires a bit of trickery anyway, because translations are not linear maps: they do not in generally send the zero vector to the zero vector as linear maps should. This difficulty is circumvented by tacking on a coordinate fixed to$~1$, which operation makes a nonzero vector out of the zero vector. In fact the matrix describes a linear operator on $\Bbb R^3$, which globally fixes the (affine) plane $z=1$, and whose restriction to that plane is a translation.
A: I want to answer my own question in response to the answer given above. In the answer Marc van Leeuwen says that ''matrix representing linear transformation should not depend on the input" 
Why exactly? What properties of linear transformations prevent that?If we find a transformation that depends on the input but satisfies $A(\alpha \textbf{v}+\beta \textbf{u})=\alpha A(\textbf{v})+\beta A(\textbf{u})$ isn't that automatically a linear transformation? 
If we try it on the above transformation we see where the problem comes from:
Let $\textbf{u} = \left( {\begin{array}{c} u^1\\u^2\end{array}}\right)$ and  $\textbf{v} = \left( {\begin{array}{c} v^1\\v^2\end{array}}\right)$ Then from the definition of $A$ we have:
$$A(\alpha \textbf{v}+\beta \textbf{u})=A\cdot \Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} \\ \alpha u^{2}+\beta v^{2}  \end{array}} \Bigg) = \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{\alpha u^2+\beta v^2}\\
\frac{y_{0}}{\alpha u^1 + \beta v^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} \\ \alpha u^{2}+\beta v^{2}  \end{array}} \Bigg) $$
$$=\Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} + x_{0} \\ \alpha u^{2}+\beta v^{2} + y_{0} \end{array}} \Bigg)$$
On the other hand 
$$\alpha A(\textbf{v})+\beta A(\textbf{u}) = \alpha \cdot \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{ u^2}\\
\frac{y_{0}}{u^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} u^{1} \\  u^{2}  \end{array}} \Bigg)+\beta \cdot \Bigg( \begin{array}{cc} 1&\frac{x_{0}}{ v^2}\\
\frac{y_{0}}{v^1} &1 \end{array} \Bigg)\cdot \Bigg( {\begin{array}{c} v^{1} \\  v^{2}  \end{array}} \Bigg)$$
$$=\Bigg( {\begin{array}{c} \alpha u^{1}+\beta v^{1} + (\alpha + \beta)x_{0} \\ \alpha u^{2}+\beta v^{2} + (\alpha + \beta)y_{0} \end{array}} \Bigg)$$
So we proved that $A(\alpha \textbf{v}+\beta \textbf{u})\neq \alpha A(\textbf{v})+\beta A(\textbf{u})$ thus the transformation provided by the matrix $A$ is not linear.
A: Another way to generate the shifts is the following.
We have
\begin{eqnarray*}
x &\rightarrow &x+x_{0} \\
y &\rightarrow &y+y_{0}
\end{eqnarray*}
Let
$$
U(a,b)=\exp [a\partial _{x}+b\partial _{y}]
$$
Then
\begin{eqnarray*}
\exp [a\partial _{x}+b\partial _{y}]x &=&\exp [a\partial _{x}]x=x+a \\
\exp [a\partial _{x}+b\partial _{y}]y &=&\exp [b\partial _{y}]y=y+b
\end{eqnarray*}
Thus the transformation is generated by $U(x_{0},y_{0})$.
