Additive Identity of a Vector Space Looking at the various axioms for vector spaces, I'm getting hung up on this one:    

Additive Identity
  The set $V$ contains an additive identity element, denoted by $0$, such that for any vector $v$ in V,     $0 + v = v$   and   $v + 0 = v$.    

It seems simple enough but in an example given -    
$V = \Bigg\{
\begin{bmatrix}
1+x\\2-x\\3+2x
\end{bmatrix}     
\Bigg| x \in \mathbb{R}\Bigg\}$    
And the additive identity is -
$0=\begin{bmatrix}
1\\2\\3
\end{bmatrix}$    
I'm confused by this as certainly, $v + 0 \neq v$.     
I'd appreciate any help in understanding this, thanks!

Your problem defines addition as follows:
$$ v_1 \oplus v_2 =
\left(\begin{array}{c}
1 + t_1 \\ 2 - t_1 \\ 3 + 2t_1 \\
\end{array}\right) 
\oplus
\left(\begin{array}{c}
1 + t_2 \\ 2 - t_2 \\ 3 + 2t_2 \\
\end{array}\right) 
=
\left(\begin{array}{c}
1 + (t_1 + t_2) \\ 2 - (t_1 + t_2) \\ 3 + (2t_1 + 2t_2) \\
\end{array}\right) 
$$
So to find your additive identity, you need to find a vector whose elements are defined such that 
$$ v_1 \oplus \text{0} =
\left(\begin{array}{c}
1 + t_1 \\ 2 - t_1 \\ 3 + 2t_1 \\
\end{array}\right) 
\oplus
\left(\begin{array}{c}
1 + t_2 \\ 2 - t_2 \\ 3 + 2t_2 \\
\end{array}\right) 
=
\left(\begin{array}{c}
1 + (t_1 + t_2) \\ 2 - (t_1 + t_2) \\ 3 + (2t_1 + 2t_2) \\
\end{array}\right) 
=
\left(\begin{array}{c}
1 + t_1 \\ 2 - t_1 \\ 3 + 2t_1 \\
\end{array}\right) 
$$
based on how $\oplus$ have been defined.
For the first element of the additive identity, you need $1 + (t_1 + t_2) = 1 + t_1$.  This will only occur of $t_2 = 0$; in other words, it will only occur if the first element of $v_2$ is $1 + 0$.  Thus, the first element of the 0 is 1.  You can solve for the others in a similar fashion.
