Is it possible to express a vector in $\mathbb{R}^2$ with a vector with $n$ components? Is it possible to express a vector in $\mathbb{R}^2$ with a vector with $n$ components?
For example, is the following true? I'm expressing vectors as a tuple, for instance $(0, 1)$ means $x=0,y=1$ and $(0,1,2)$ means $x=0,y=1,z=2$
$$(a, b) = (a, b, 0) = (a, b, 0, 0) ...$$
I.e., is it okay to take a 3D vector in the form of $(x, y, 0)$ and put it into a 2 dimensional space in the form of $(x,y)$?  
Can we represent any dimension like this, with a vector from a higher dimension as long as the higher dimension components are zero?
 A: They are different objects. In fact they are in different spaces. However, there is an isomorphism between the subspaces. For example, if $W=\{ (x,y,0,0) : x,y \in \mathbb{R} \} \subset \mathbb{R}^4$, then the function $f : \mathbb{R}^2 \to W, f((x,y)) = (x,y,0,0)$ is an isomorphism. This means it is bijective, it is linear, and its inverse is linear. It is only in this sense that $W$ and $\mathbb{R}^2$ are "the same".
A: If you write a vector $v_n$ from $\mathbb{R}^n$ as a vector $v_m$ in $\mathbb{R}^m$ where $m > n$ they are not "equal" in the strict sense of the word. $v_m$ is a higher dimensional copy of $v_n$, but $v_n \neq v_m$. You could easily define a linear transformation T by $Tv_n = v_m$. If you applied this linear transformation to all vectors in $\mathbb{R}^n$ then you would get a "copy" of $\mathbb{R}^n$ that's embedded in $\mathbb{R}^m$. But it's probably most important to note that you could write all $(x,y) \in \mathbb{R}^2$ as either $(x, y, 0) \in \mathbb{R}^3$, $(x, 0, y) \in \mathbb{R}^3$, or $(0, x, y) \in \mathbb{R}^3$. All of these would give you "copies" of $\mathbb{R}^2$ embedded in $\mathbb{R}^3$ but clearly these three sets of vectors are not equal. Hopefully this gives you a bit more intuition about why a vector from a lower dimension is not the same as its higher-dimensional analogue. 
