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I'm trying to learn a little about vector spaces and such before heading off to university after the Summer. My question is a misunderstanding I have over, what seems, two conflicting pieces of information:

I read that the vector space $\Bbb{R}^2$ is not a sub space of the vector space $\Bbb{R}^3$.

I don't understand why this is, given the case that if you consider the set of all vectors:

$S=\begin{bmatrix}x\\y\\0\end{bmatrix}$

which is by definition a vector space because it follows the conditions of being closed under addition and scalar multiplication. All of $S$ can be found in the vector space $\Bbb{R}^3$, and therefore it is a sub space of $\Bbb{R}^3$.

However, isn't $S$ nothing other than the real $x$-$y$ plane, which is, by definition the vector space $\Bbb{R}^2$? This would suggest that $\Bbb{R}^2$ is a sub space of $\Bbb{R}^3$. But then we look again at the first statement, which says that this cannot be the case! Please show me the flaw in my thinking.

Thank you all in advance!

Sourav Ghosh
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Adam Gardner
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    Technically, we say that $\mathbb R^2$ is isomorphic to a subspace of $\mathbb R^3$. If your instructor differentiates between the two, you may want to ask if $\mathbb R\subset\mathbb C$. – John Douma Jul 26 '22 at 14:20
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    $\Bbb R^2\times 0$ is contained in $\Bbb R^3=\Bbb R\times \Bbb R\times \Bbb R$, to be formally correct. Of course $\Bbb R^2\times 0\cong \Bbb R^2$ as vector spaces. – Dietrich Burde Jul 26 '22 at 14:20
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    The vectors with third component $0$ form a copy of $\mathbb{R}^2$ in $\mathbb{R}^3$. Those vectors are not pairs of numbers, so they are not actually in $\mathbb{R}^3$. – Ethan Bolker Jul 26 '22 at 14:21
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    There is a subspace of $\Bbb R^3$ which is isomorphic to $\Bbb R^2$... that is to say it is "the same" as $\Bbb R^2$ in essentially every way that matters except for how it is formally defined and as such they are not equal. – JMoravitz Jul 26 '22 at 14:21
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    Re: "You may want to ask if $\Bbb R\subset \Bbb C$" In maths we do very frequently blur the lines between certain objects. If I talk about "the number 1" I could freely switch between talking about the natural number $1$, the integer $1$, the rational number $1$, the real number $1$, the complex number $1$ and so on depending on context. The formal definitions of each of these objects is different so from a set theoretical standpoint they are not "equal", but they all do share some property of "one-ness" that even though they may not be "equal" they are "essentially equal" – JMoravitz Jul 26 '22 at 14:24
  • It's really just a technicality meant to emphasize that adding a dimension is not a trivial operation, even if the added dimension is set to zero. – zcsttn Jul 26 '22 at 14:25
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    Wait, so $\mathbb R$ is not a subset of $\mathbb C$? – insipidintegrator Jul 26 '22 at 14:30
  • @insipidintegrator Have a look here: https://math.stackexchange.com/questions/1553537/why-mathbbr-is-a-subset-of-mathbbc - from the looks of things, just as R2 is not a sub space of R3, R is not a subset of C! – Adam Gardner Jul 26 '22 at 14:35
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    Answer depends on definition what is $\mathbb{C}$. Usually $\mathbb{C}={x+iy|x,y\in\mathbb{R}}$. In this case $\mathbb{R}$ is subset of $\mathbb{C}$ like set of all $S$-vectors from the question is subset of $\mathbb{R}^3$. – Ivan Kaznacheyeu Jul 26 '22 at 14:35
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    @IvanKaznacheyeu It's the same issue. Is $(1,0)\in\mathbb C=1\in\mathbb R$? This is precisely the example that Adam presents with column vectors. – John Douma Jul 26 '22 at 16:16

2 Answers2

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$T:\Bbb{R^2}\to\Bbb{R}^3$ defined by $$T(x, y) =(x, y, 0) $$

is a one-to-one linear map.

$\Bbb{R}^2\cong \text{im} T \subset \Bbb{R}^3$

I.e $\Bbb{R}^3$ contains a isomorphic copy of $\Bbb{R}^2$ as subspace. Infact that subspace is the $xy$-plane.

Not exact $\Bbb{R}^2$ but an isomorphic copy of $\Bbb{R}^2$ which have same linear structure as of $\Bbb{R}^2$ but as a set they are different. Since $(x, y) \neq (x, y, 0) $ . Infact we can't compare them as they have different length!

Sourav Ghosh
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It depends on who you ask. Yes, $S=\{(x,y,0)\,|\,x,y\in\mathbb{R}\}$ is a two-dimensional subspace of $\mathbb{R}^3$. However, technically, $S\neq\mathbb{R}^2$, because the lengths of the vectors are not the same.

In $S$, you have length-3 vectors with the final component set to zero; in $\mathbb{R}^2$, we just have length-2 vectors. So, the mathematical statement $S=\mathbb{R}^2$ which means $(\forall s\in S) s\in \mathbb{R}^2$ and $(\forall z\in\mathbb{R}^2) z\in S$ does not hold, since a statement like $(1,1,0)\in\mathbb{R}^2=\{(x,y)\,|\,x,y\in\mathbb{R}\}$ is false (due to the third component in $(1,1,0)$.)

Zim
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