Why a real Vector Space has either one or infinitely many vectors? I have a hard time understanding the above question, so could someone help?
 A: if $v \in V$ is a vector in the vector space, then for every $\lambda \in \mathbb R$ also $\lambda v \in V$ (This is an axiom)
So either is $v = 0$ is the only vector in the vector space,  so all the vectors in the form $\lambda v$ are equal to $0$, or there are infinitel many, depending on $\lambda$
What I mean is: if for example $v = (1,1)$ is in out space, then also $2v = (2,2)$ has to be in the vector space (because of our axiom). But then also $3v = (3,3)$ and so on, for every $\lambda$: $(\lambda, \lambda)$ has to be in $V$
A: $\mathbb{R}^2$ is a good example of vector space. It's a plane. if you draw two axes: $x$ and $y$ you get an $\mathbb{R}^2$ space. You can have infinitely many points in this space.
For example the vector $v=(1,1)$. Also by definition of vector space any multiplication of $v$ lies in the vector space. In our case it's any point lying on the line defined by the points $(0,0)$ and $(1,1)$. E.g.: $(2,2)$, $(\pi,\pi)$, $(10^{100},10^{100})$...
A: If you real vector $E$ space is no $\{0\},$ then it contains a vector $x\in E\setminus\{0\}.$ So you can consider the (no $\{0\}$) subspace $\mathbb{R} x=\{\lambda\cdot x\,;\,\lambda\in\mathbb{R}\}.$ Note that it is a  $1-$dimensional vector space and then you can make an isomorphism between $\mathbb{R}x$ and $\mathbb{R}^1=\mathbb{R}.$ In particular, it is a bijection and, as $\mathbb{R}$ has an infinity of elements, it is the case for $E$ too.
