Confused about the definition of subspace

Question

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?

Answer 1

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.

Answer 2

$\mathbb{R}^2$ 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 $\mathbb{R}^2$.

Answer 3

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.

Answer 4

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.

Answer 5

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 $\mathbb{R}^3$.