Is the vector $(1,2,0)$ the same as $(1,2)$? Say I have a vector in $\mathbb{R^3}$ and a linear mapping 
$$ f: \mathbb{R^3} \rightarrow \mathbb{R^3}, (x,y,z) \mapsto (x,y,0) $$
then the $im(f) \subseteq \mathbb{R^3}$ will be a 2 dimensional vector space, correct? 
This is where the question comes in, isn't the vector $(1,2,0) \in im(f)$ necessarily two dimensional ( since $im(f)$ is) and therefore equal to $(1,2)$? Does that imply every vector of the form $(1,2,0,0,0,0)$ is equal to $(1,2)$? I know their ranks are equal, but am not sure, if this is the same as being equal.
 A: These vectors are not the same, becasue one is an element of $\mathbb{R}^3$ and the other is an element of $\mathbb{R}^2$.
However, we can think of each plane in $\mathbb{R}^3$ as a copy of $\mathbb{R}^2$ (they are isomorphic). And therefore think of $(1,2,0)$ as $(1,2)$ if it is clear from the context.
On the other hand, there is no unique way to embed $\mathbb{R}^2$ in $\mathbb{R}^3$, as there are infinite planes in $\mathbb{R}^3$.
For example if your function was defined by $f(x,y,z)=(x,y,1)$ it would be convenient to identify $\mathbb{R} \times \mathbb{R}\times \{1\}$ with $\mathbb{R}^2$.
A: The vector $(1,2,0)$ is still three-dimensional; it's just that the third component happens to be $0$. As a test, notice that the vector sum $(1,2,0)+(1,1,1) = (2,3,1)$ makes sense, whereas $(1,2)+(1,1,1)$ does not.
There is an obvious relationship between vectors of the form $(x,y,0)$ and those of the form $(x,y)$; that is, the $xy$-plane in $\mathbb R^3$ looks like $\mathbb R^2$. So they can be mapped to each other in a natural way, but they are not the same.
A: If $S$ is a subspace of $\Bbb R^n$, then $\dim S = 2$ just means that $S$ has a basis of cardinality $2$. And no, $(1,2)$ is not the same as $(1,2,0)$ because $(1,2) \in \Bbb R^2$ and $(1,2,0) \in \Bbb R^3$. Though in a sense, $A:= \{(x,y,0): x,y \in \Bbb R\}$ is "the same" as $\Bbb R^2$ because the map $T: A \to \Bbb R^2$, $(x,y,0) \mapsto (x,y)$ is an isomorphism (linear and bicontinuous).
