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Consider the vector $\vec{v} = (p, (x_1, y_1))$, where $p \in \mathbb{R}^2 = (p_1, p_2)$ is a coordinate point that denotes where the tail of the vector is, and $(x_1, y_1) \in \mathbb{R}^2$ denotes the vector's head.

The other day, I had a discussion the in which these two points were being defended:

  • The vector $\vec{v}~' = (p~', (x_1, y_1, 0))$, where $p \in \mathbb{R}^3 = (p_1, p_2, 0)$, is a different vector than $\vec{v}$. Upon viewing it from a third dimension—you change the vector. (A vector that exists only in two dimensions, and a vector, with similar components, that exists in three dimensions is a different vector with different properties).
  • The vector $\vec{v}~'$ is the same vector as $\vec{v}$, you're only changing the space from which you view and analyze it. (Something like: The vector already exists in 3 [and higher] dimensions; prior, it was being looked at only through $2D$ goggles, so to speak.)

My question is: Is one of these accepted as true, is it a current issue of contention (like "math is discovered" and "math is invented" or something similar), or is it just a semantic issue with no real relevance?

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  • $\begingroup$ But then, if I have another vector $\vec {v''} = ((0,0), (x_1-p_1, y_2-p_2))$, is this the same vector $\vec{v}$ viewed differently? What about $\vec {v'''} = ((0,0), (-y_2+p_2,x_1-p_1))$? $\endgroup$ – peterwhy Jun 16 '17 at 18:13
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From a mathematical point of view the answer is simply that a vector in $\mathbb{R}^2$ is a different thing from a vector in $\mathbb{R}^3$ and that $\mathbb{R}^3$ contains infinetly many subspace that are isomorphic to $\mathbb{R}^2$ ( ''copies'' or $\mathbb{R}^2$), but these subspace are different things wen viewed in $\mathbb{R}^3$.

So, has this some ''philosophycal'' consequence about the ''real'' existence of some one of these vectors? I'm not a philosopher, but I suspect that this has something to do with the question about the dimension of the space in which our ''real world'' lives.

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  • $\begingroup$ I'm partial to this answer (but that's because it was the point I was arguing!) I don't want to accept it yet—Aniruddha's answer seems reasonable, as well. What, I think, would solve this is: does $\mathbb{R}^2$ being a subspace of $\mathbb{R}^3$ just mean $\mathbb{R}^3$ has a subspace that is isomorphic to $\mathbb{R}^2$? (In which case, the first point [and you] are correct). On the other hand, as Aniruddha's answer seems to imply something different. $\endgroup$ – AmagicalFishy Jun 16 '17 at 18:57
  • $\begingroup$ Definitely $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$. See, eg : math.stackexchange.com/questions/1247108/… $\endgroup$ – Emilio Novati Jun 16 '17 at 19:03
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The second point appears more appropriate. Since, $\mathbb{R^2}$ is a subspace of $\mathbb{R^3}$, the vector was not changed. In fact, the first statement gives that you are observing in 2 - D (specifically in XY Plane) and then in the second statement, you observe from 3 - D (specifically the XYZ space). Therefore, the vector remains the same, however, your observation will change! Since you view it from a space (rather than a plane), you will get to observe one more property of that vector that it has no "height" or the z - component.

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    $\begingroup$ $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$! $\endgroup$ – AmagicalFishy Jun 16 '17 at 19:27

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