Vector subspaces I had a doubt while studying Linear Algebra. I read that for vector space $ R^3, R^2 $ cannot be termed a subspace because essentially there is no translation between the $ v = (x,y,z) \in R^3 $ and $ w = (x,y) \in R^2 $. 
Now my question is there are cases when the dimension of a subspace is smaller than the dimension of the vector space. If that be the case then why can $ R^2 $ not be a subspace of $ R^3 $ since it does fulfill the properties of a subspace. Can you give an example of subspace of $ R^3 $ which has dimension less than 3.
 A: A subspace of $\mathbb{R}^3$ is made up of elements of $\mathbb{R}^3$, i.e. ordered triples of number like $(x,y,z)$.
On the other hand, $\mathbb{R}^2$ is made up of order pairs of number, e.g. $(u,v)$.
A vector in $\mathbb{R}^2$ (an ordered pair) cannot be an element of $\mathbb{R}^3$ because it's not an ordered triple.
We can identify $\mathbb{R}^2$ with the subspace $\{(x,y,0) : x,y, \in \mathbb{R}\}$ of $\mathbb{R}^3$, but that's all.
It's much the same with $\mathbb{R}$ and $\mathbb{R}^2$. Consider the number $1$ in $\mathbb{R}$. We cannot think of $1$ as an element of the plane $\mathbb{R}^2$; it doesn't make sense to write $1 \in \mathbb{R}^2$. However, we may identify $1 \in \mathbb{R}$ with $(1,0) \in \mathbb{R}^2$. That way $\varphi : \mathbb{R} \hookrightarrow \mathbb{R}^2$, where $\varphi(x) := (x,0)$. However, $\mathbb{R} \not\subset \mathbb{R}^2$.
A: One of the properties of a subspace $S$ of a vector space $V$ is that $S\subset V$, in other words that all elements of $S$ are elements of $V$. For $S=\Bbb R^2$ and $V=\Bbb R^3$ this is false, since pairs of numbers are not triples. So $\Bbb R^2$ does not fulfill the properties of a subspace of $\Bbb R^3$
On the other hand $\{\, (x,y,z)\in\Bbb R^3\mid x+y=0\,\}$ is a subspace of $ \Bbb R^3$, and its dimension is less than$~3$. You can replace the linear equation in the formula by other ones, like $x+2y-z=0$ or $y=0$ and get other examples.
