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Consider $ v_1 $ and $ v_2 $:

$ \{v_1 \in \mathbb{R}^m\mid v_1 = (x_1,x_2,...,x_{m})\}\\ \{v_2 \in \mathbb{R}^{m+1}\mid v_2 = (x_1,x_2,...,x_m,0)\} $

Is $v_1 = v_2$ even they're belonging to different spaces or in this case, $ \mathbb{R}^m = \mathbb{R}^{m+1}$ when the last coordinate is 0?

Augusto Ferreira
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    They are clearly different objects. But we often regard one vector space as embedded in another, which allows us to identify objects in one with objects in the other in a natural way. – almagest Apr 08 '16 at 09:32
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    I wonder, why not $(x_1,x_2,\ldots,x_m)=(0,x_1,x_2,\ldots,x_m)$ or even $(x_1,x_2,\ldots,x_m)=(x_1,x_2,0,x_3,\ldots,x_m)$? These embeddings of $\mathbb R^m$ in $\mathbb R^{m+1}$ rarely if ever are mentioned, but is this for any reason of mathematical "substance" or is it just a matter of preference and style? – David K Apr 08 '16 at 13:31

2 Answers2

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Strictly speaking, $(v_1,\ldots,v_m,0)\ne (v_1,\ldots,v_m)$. Then again, the obvious injective map $\iota\colon\Bbb R^m\to \Bbb R^{m+1}$ is sometimes viewed as being so natural that we sometimes identify the image $\iota(\Bbb R^m)$ and $\Bbb R^m$ and say, for example $\Bbb R^m\subseteq \Bbb R^{m+1}$ (instead of more correctly $\iota(\Bbb R^m)\subseteq \Bbb R^{m+1}$). In such a context, we would also identify $(v_1,\ldots,v_m)$ with $\iota(v_1,\ldots,v_m)=(v_1,\ldots,v_m,0)$ and say they are equal.

There are contexts where this comes really natural, such as when whe construct, e.g., rationals as equivalence classes of pairs of integers and yet view the integers as subset of the rationals; every step of $\Bbb N\subset\Bbb Z\subset \Bbb Q\subset \Bbb R\subset \Bbb C$ is of this kind: We construct the larger object from the smaller and identify the original set with its different-looking copy in the larger set. However, for the case at hand $\Bbb R^m\to\Bbb R^{m+1}$, the embedding may be what first comes to mind - but it is not really natural. Hence here I would rather not make the identification (unless after specifically mentioning the choice of embedding).

As a sidenote, Let $V$ be any vector space, $v\in V$ any element, and $\alpha$ any object with $\alpha\notin V$ (for example, my neighbour's dog). Then we can define a vector space $V'$ containing $\alpha$ in place of $v$. This way, we can easily construct vector spaces having arbitrary elements in common (and in fact the same elements playing different "roles" in the different vector spaces)

Hagen von Eitzen
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  • I like this answer. Could you say that for subspaces $U_{1}$ and $U_{2}$ of $V$ with bases $B_{1}$ and $B_{2}$ respectively, then any vector in $\text{Span}\left( B_{1} \cap B_{2} \right)$ is in $U_{1}$ and in $U_{2}$ ? I could then easily make different vector spaces $U_{1}$ and $U_{2}$ which are different yet have non-trivial intersection. – Shai Apr 08 '16 at 09:59
  • One can get "arbitrary elements in common" more easily: ​ Just consider a space with dimension greater than 1, and strict subspaces containing any given vector. ​ ​ ​ ​ –  Apr 08 '16 at 12:37
  • You can speak of canonically isomorphic, for instance? – djechlin Apr 08 '16 at 14:13
  • As for $\mathbb N\subset\mathbb Z\subset\dots$ --- I've seen constructions of $\mathbb Z$ when the last step is to forcefully replace $\iota(\mathbb N)$ with $\mathbb N$ itself, and to fix all the introduced relations/operations/functions. – Joker_vD Apr 08 '16 at 14:18
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    @Joker_vD Indeed, that's what one can do at any of the steps, introducing complicated exceptions to all definition. The mere knowledge that on can do should convince one that one shouldn't do it as explicitly, and recall that the constructions just show existence - rather let $\Bbb Z$ be any unitary ring that has exactly one hom into every unitary ring, and let $\Bbb R$ be any completely ordered archimidean field, etc. This way, whatever your $\Bbb C$ is, you'll find exactly one $\Bbb R$ in it, exactly one $\Bbb Q$ in that, exactly one $\Bbb Z$ in that, exactly one $\Bbb N$ in that. – Hagen von Eitzen Apr 08 '16 at 15:12
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The presumption that two different vector spaces can contain the same vector is correct. For instance, if we consider distinct, though overlapping, subspaces of the same vector space then they certainly contain a common vector. For instance in $\mathbb{R}^3$ the $xy$-plane and the $x$-axis have all their vectors in common, yet they are distinct subspaces. I suppose it depends on what conditions you would like your subspaces to satisfy.

In your example, I would not say the vectors are the same since the space in which we view the vector is important. I would argue instead that

$$ v_2 \;\; =\;\; \left [ \begin{array}{cc} v_1 & 0 \\ \end{array} \right ]. $$

They are inherently different since they are objects of different dimensions, even though they contain "nearly" the same information.

Mnifldz
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