Translations in two dimensions - Group theory I have just started learning Lie Groups and Algebra. Considering a flat 2-d plane if we want to translate a point from $(x,y)$ to $(x+a,y+b)$ then can we write it as :
$$   \left( \begin{array}{ccc}
x+a  \\
y+a  \end{array} \right) =
  \left( \begin{array}{ccc}
x  \\
y  \end{array} \right) + 
  \left( \begin{array}{ccc}
a  \\
b  \end{array} \right)$$
Now the set of all translations  $ T =   \left( \begin{array}{ccc}
a  \\
b  \end{array} \right) $ form a two parameter lie group (I presume) with addition of column as the composition rule.
If that is so, how do I go about finding the generators of this transformation.
I know the generators of translation are linear momenta in the corresponding directions. But I am not able to see this here.
PS:  In my course I have been taught that the generators are found by calculating the Taylor expansion of the group element about the Identity of the group. For instance, $\operatorname{SO}(2)$ group 
$$ M =   \left( \begin{array}{cc}
\cos \:\phi & -\sin \:\phi \\
\sin \:\phi &  \cos \:\phi \end{array} \right) $$
I obtain the generator by taking 
$$ \frac{\partial M}{\partial \phi}\Bigg|_{\phi=0} = 
 \left( \begin{array}{cc}
0  & -1 \\
1 &  0 \end{array} \right) $$
Now if I exponentiate this, I can obtain back the group element. My question how do I do this for Translation group. 
EDIT :This edit is to summarise and get a view of the answers obtained.
Firstly, the vector representation of the translation group (for 2D) would in general have the form : 
$$ \begin{pmatrix} 1 & 0 & a_x\\ 0 & 1 & a_y \\ 0 & 0 & 1 \end{pmatrix}\ $$
with generators (elements of Lie algebra) 
$$  T_x =\begin{pmatrix} 0 & 0 & i\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ , \;\; 
T_y = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & i \\ 0 & 0 & 0 \end{pmatrix}\  $$
Secondly, the scalar-field representation of the same is given by the differential operators
$$ exp^{ i(a_x\frac{\partial}{\partial x}+ a_y\frac{\partial}{\partial y} )} $$
with generators 
$$ T_x^s = i\frac{\partial}{\partial x},\;\;T_y^s = i\frac{\partial}{\partial y} $$
The Lie algebra is two-dimensional and abelian : $ [T_x,T_y] = 0$ 
 A: The issue is that translations add an inhomogeneous piece and so there is no matrix associated with it. Change it to the following so that we can associate a matrix:
$$
\begin{pmatrix} x' \\ y' \\ 1 \end{pmatrix}=
\begin{pmatrix} 1 & 0 & a\\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\ 
\begin{pmatrix} x \\ y \\ 1 \end{pmatrix}\ .
$$
Note that $(x,y,1)$ has the same data as $(x,y)$ and thus both are realizations of Euclidean space. Now you can write the $3\times 3$ matrix as the exponential of (the nilpotent matrix)
$$
\begin{pmatrix} 0 & 0 &a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{pmatrix}\ .
$$
This construction is related to the fact that the two-dimensional Euclidean group can be obtained as a (Inonu-Wigner) contraction of $SO(3)$ (but don't worry if this statement doesn't make sense right away). So you now obtain three generators for the Lie algebra for the Euclidean group:
$$
P_x\sim \begin{pmatrix} 0 & 0 &1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\quad,\quad
P_y\sim \begin{pmatrix} 0 & 0 &0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}\quad\textrm{and} \quad M \sim \begin{pmatrix} 0 & 1 &0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}
$$
A: For one thing, it's impossible to write a nonzero translation on $\mathbb R^2$ as a linear transformation (matrix).  I encourage you to try to prove this.  Therefore, since exponentiating a matrix gives another matrix, given a nonzero translation $T(\mathbf x) = \mathbf x + \mathbf a$, there does not exist a matrix $M$ for which $T(\mathbf x) = e^M\mathbf x$ for all $\mathbf x\in\mathbb R^2$.
However, let $f$ denote a smooth real-valued function on the real line, then translations act on such a function as follows:
\begin{align}
  (T_af)(x) = f(x+a).
\end{align}
Now, notice that Taylor expansion gives
\begin{align}
  f(x+a) 
  &= \left(1 +a\frac{d}{dx}  + \frac{a^2}{2!}\frac{d^2}{dx^2} + \cdots + \frac{a^n}{n!}\frac{d^n}{dx^n} + \cdots\right)f(x) \\
  &= \exp\left(a\frac{d}{dx}\right) f(x)
\end{align}
to that the translation operator can be written as the exponential of the derivative;
\begin{align}
  T_a = \exp\left(a\frac{d}{dx}\right)
\end{align}
But recall that, up to normalization, the derivative is precisely the momentum operator in the position space representation of a particle moving on the real line in quantum mechanics.  This is one way of seeing that momentum generates translations.
This can easily be generalized to higher dimensions where the generator of translations in the direction of the standard ordered basis vector $\mathbf e_1$ is $\partial/\partial x^i$.
A: As suresh mentioned if the vector is just a two component object then you can't translate it without expanding the vector. However, if you consider the vector to be variable (which are essentially infinite vectors) then it can be translated. 
To find the differential form of a translation, start with the translation of a 1D dimensional vector, $x$:
\begin{align} 
e ^{ i\epsilon {\cal P} } x & = x + \epsilon \\ 
\left( 1 + i \epsilon {\cal P} \right) x &= x + \epsilon \\ 
{\cal P} x & = - i  
\end{align} 
Thus we must have $ {\cal P} =  - i \frac{ \partial }{ \partial x } $. 
Now it is easy to extend this to two dimensions:
\begin{align} 
e ^{ i\epsilon _x  {\cal P} _x + i \epsilon _y {\cal P} _y  } \left( \begin{array}{c} 
x \\  
y 
\end{array} \right)
& = \left( \begin{array}{c} 
x + \epsilon _x  \\  
y + \epsilon _y  
\end{array} \right) \\
i \left( \epsilon _x {\cal P} _x + \epsilon _y {\cal P} _y \right) \left( \begin{array}{c} 
x \\  
y 
\end{array} \right) &= \left( \begin{array}{c} 
x + \epsilon _x  \\  
y + \epsilon _y  
\end{array} \right) 
\end{align} 
where we have two different generators since you have two degrees of freedom in the transformation you gave in your question. This expression requires,
\begin{align} 
& {\cal P} _x = \left( \begin{array}{cc} 
- i \frac{ \partial }{ \partial x } & 0 \\  
0 & 0
\end{array} \right) \\ 
& {\cal P} _y = \left( \begin{array}{cc} 
0 & 0 \\  
0 & - i \frac{ \partial }{ \partial y } 
\end{array} \right)
\end{align}
