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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.

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    The short answer is "No". They are different things, and they live in different spaces. This is not like the number $2.5$ being the same as $2.50$. – 2017-01-20
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    Technically they are not "equal", since they are not elements of the same set. You are right that the image of $f$ is isomorphic to $\Bbb{R}^2$. Rather than saying "equal", it would be more correct to say that $(1,2,0)$ is associated to $(1,2)$ under the isomorphism. – 2017-01-20
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    What is your definition of equal? Depending on that people may refer to them as "equal", you might also want to think about: What are criteria for distinguishing objects. What are its characteristic properties? What does make it unique? Thinking about that might give you more insight, than just a aplain answer here. – 2017-01-20
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    I think $(1,2)$ is an element of $(1,2,0)$ within the rigorous set theory definition of ordered pair. – 2017-01-20

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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 $\mathbb{R}^2$.

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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.

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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).