# Confused about the definition of subspace

The definition from my textbook is:

A subspace of a vector space is a set of vectors that satisfies two requirements:

If $v$ and $w$ are vectors in the subspace and $c$ is any scalar, then

(1) $v + w$ is in the subspace.

(2) $cv$ is in the subspace.

And my textbook says vector space $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$ but let say $V=\begin{pmatrix}a\\b\end{pmatrix}$ and $W=\begin{pmatrix}c\\d\end{pmatrix}$, so $V$ and $W$ are all in $\mathbb{R}^2$, and clearly $V+W$ is in $\mathbb{R}^2$, $cV$ or $cW$ is in $\mathbb{R}^2$ too, so the first two requirements are met, why we say $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$? It looks like we have to add $V$ and $W$ should be in $\mathbb{R}^3$ as well, but why we don't have this in the definition?

• How on earth is $\mathbb{R}^2$ not a subspace of $\mathbb{R}^3$? Surely there's something wrong, as you have said. Or maybe it's better to say that $\mathbb{R}^2$ is isomorphic to a subspace of $\mathbb{R}^3$, but that's all the same to me. – астон вілла олоф мэллбэрг Apr 11 '16 at 13:04
• The question is, whether $V$ is an element of $\Bbb R^3$. Is it? If you identify it with $(a,b,0)^T$ you can understand it as an element of $\Bbb R^3$ but that's not given. – Peter Franek Apr 11 '16 at 13:04
• Neither of those vectors exist in $R^3$. There exist no vectors in $R^2$ that exist in $R^3$. What you showed was that $R^2$ is a subspace of $R^2$. – J. Bush Apr 11 '16 at 13:06
• @whoisit Note that a vector subspace is by definition a subset that additionally satisfies some properties. So, if $V$ is in a subspace of $\Bbb R^3$ then it should also be in $\Bbb R^3$. – Peter Franek Apr 11 '16 at 13:16
• @астонвіллаолофмэллбэрг This can lead to all sorts of problems. There are MANY subspaces of $\mathbb{R}^{3}$ isomorphic to $\mathbb{R}^{2}$, so if you say "take the subspace $\mathbb{R}^{2}$ of $\mathbb{R}^{3}$", it is not at all clear which one you mean. – Morgan Rodgers Apr 11 '16 at 13:29

Your definition misses the crucial point that the subspace must be a subset of the parent space. So in particular every vector in the subspace must also be a vector in the parent space.

$\begin{pmatrix}a\\b\end{pmatrix}$ is not an element of $ℝ^3$, because it has two components and vectors in $ℝ^3$ have three.

• In fairness, the definition the OP quotes does not make this point very clearly -- it's hidden in the phrase "a set of vectors" where it must be implicit that "vectors" means "members of the only vector space we have mentioned recently". – Henning Makholm Apr 11 '16 at 16:33

$\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$, but it can be canonically identified with a subspace. Many mathematicians identify $\mathbb{R}^2$ with $$\left\{ \begin{pmatrix} v_1 \\ v_2 \\ 0 \end{pmatrix} \mid v_1, v_2 \in \mathbb{R} \right\}.$$ As such, $\mathbb{R}^2$ is a subspace of $\mathbb{R}^3$. More precisely, we should say that $\mathbb{R}^3$ contains, as a vector subspace, a copy of $\mathbb{R}^2$.

• the main point of my question is that "do we need to add $V$ and $W$ should be in the main vector space too, not just in the subspace. Because if $V$ and $W$ doesn't necessarily in $R^3$, then $R^2$ should be a subspace of $R^3$, isn't it? – whoisit Apr 11 '16 at 13:12
• @whoisit The point here is that $\mathbb{R}^2$ is not a subset of $\mathbb{R}^3$. A subspace must be first of all a subset. – Siminore Apr 11 '16 at 14:17
• "Everybody tends to" it not quite what the term "canonical" seeks to convey (though admittedly opinions about what it does mean diverge). IMO canonical is used when there is some (nice) property that sets apart one choice among all possible choices. For instance the standard basis of $\def\R{\Bbb R}\R^n$ is a canonical choice, since it is the only one for which coordinates of any vector are given by the vector components itself. There is no clear such property that sets apart this way of embedding $\R^2\to\R^3$ (though one could make one up). I would just call this the most obvious embedding. – Marc van Leeuwen Apr 11 '16 at 15:47
• My favorite mapping is $(v_1,v_2) \to (v_1, v_2, 1)$, as with this you can perform affine transformations in $\mathbb{R}^2$ by performing linear transformations in $\mathbb{R}^3$. – Dan Bryant Apr 11 '16 at 21:21

I think you are thinking of the $x-y$ plane as a part of $x-y-z$ space, but that's not the right way to think abstractly about $\mathbb{R}^2$ and $\mathbb{R}^3$ .

Vectors in $\mathbb{R}^2$ have two components, not three, so $\mathbb{R}^2$ is not a subset of $\mathbb{R}^3$ so it can't be a subspace.

Your definition of a subspace of a vector space is fine. However, there is an important distinction to make between $\mathbb{R}^2$ and $\mathbb{R}^3$. If ${\bf v}\in\mathbb{R}^3$ then we can write ${\bf v}=(v_1,v_2,v_3)$. We notice that this vector has three components. The last component can be zero, giving a vector ${\bf v'}=(v_1,v_2,0)$, and we note that this defines a point in the $xy$ plane, but ${\bf v'}\in\mathbb{R}^3$ still. If ${\bf u}\in\mathbb{R}^2$, then we can write ${\bf u}=(u_1,u_2)$. This vector has two components, rather than three.

If ${\bf u}=(u_1,u_2,0)$ and ${\bf u'}=(u_1,u_2)$ are vectors in $\mathbb{R}^3$ and $\mathbb{R}^2$, you must realise that ${\bf u}\neq {\bf u'}$. Thus, if ${\bf u}$ is in $\mathbb{R}^2$, then it is not in $\mathbb{R}^3$, since it has two components, rather than three.

• I want both $V$ and $W$ to have only two components. Do we need to modify the definition to be "$V$ and $W$ should also be in $R^3$, not just only in $R^2$? because if we leave it out, $R^2$ can be a subspace of $R^3$, even there are only two components in V and W – whoisit Apr 11 '16 at 13:16
• I'm unsure what you mean. But if your vectors only have two components then they are not elements of $\mathbb{R}^3$. – user230944 Apr 12 '16 at 3:47

You need to avoid abuse of notation. For most considerations, $\mathbb{R}^2$ is not a subspace of $\mathbb{R}^3$ nor is it $\mathbb{R}^2 \subset \mathbb{R}^3$, they don't even contain the same components ($\mathbb{R}^2$ has 2 and $\mathbb{R}^3$ has 3, thus it cannot be a subset of the other). Rather though... it is $\textit{isomorphic}$ to a subspace in $\mathbb{R}^3$, mainly:

$$\begin{pmatrix}x\\y\\0\end{pmatrix}, x, y \in \mathbb{R}$$

That is the $x -y$ plane in the $x - y - z$ space of $\mathbb{R}^3$.