Confused about the definition of subspace

Solution 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.

Solution 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 $\mathbb{R}^2$.

Solution 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.

Solution 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.