Dimensions of a basis of a coordinate space I need a little clarification on the relationship between the basis, its dimension and their corresponding real coordinate space.
Suppose we are operating in the fourth coordinate space $\mathbb{R}^4$. I know the basis of a coordinate space should span the space and consist of linearly independent vectors. So for $\mathbb{R}^4$, the following is a suitable basis:
Basis $= \{\{1,0,0,0\},\{0,1,0,0\},\{0,0,1,0\},\{0,0,0,1\} \}$. This basis has a dimension of 4. Up to this point, I'm fine. I get confused when we start lowering the dimension. Does reducing the dimension of the basis mean that it no longer spans $\mathbb{R}^4$? Instead, it spans a subspace of $\mathbb{R}^4$ because there are now vectors in the coordinate space that are no longer represented when the dimension is reduced? Any extra pointers on this topic would also be appreciated.
Also, what would then be the difference between three-dimensional vectors in $\mathbb{R}^4$ and vectors in $\mathbb{R}^3$. I realize that the height of their columns would be different, but what is the significance of that?
 A: Let's reduce the dimesion by two to see what's going on: take the vector $(1,2)$ in $\mathbb R^{2}$. It spans a subspace given by $S=\left \{ a(1,2):a\in \mathbb R \right \}$ This is simply the line $y=2x$. Indeed, the one-dimensional subspaces of $R^{2}$ are lines through the origin. That is, "copies" of $R$ sitting in $R^{2}$.
So to address your question, if we take the vectors $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$ in $\mathbb R^{4}$, we see that they span a subspace $S=\left \{ a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0):a,b,c\in \mathbb R \right \}$ which is simply $S=\left \{ a,b,c,0):a,b,c\in \mathbb R \right \}$. Now because $a,b$ and $c$ can be any triple of real numbers, we see that we have a "copy" of $R^{3}$ sitting in $R^{4}$.
In general, in a vector space of dimesion $n$ the vector subspaces of dimension $k<n$ are copies of $k-$dimensional spaces that "live" in the big space. 
A: You're speaking of the "dimension of a basis" which makes little sense: a basis is a set, not a space; we don't talk about its dimension, it's not defined. However, it is true that the cardinality (number of elements) of a basis that spans a certain vector space is equal to the dimension of said space. 
As for your first question, it's as you say: if you remove a vector from that set, it won't span $\mathbb{R}^4$ but a subspace of it.
As for your second question, 3D vectors don't exist in  $\mathbb{R}^4$ simply because $\mathbb{R}^4$ is defined as the set that contains all four dimensional vectors (and only them). You can, of course, pretend you're working on  $\mathbb{R}^3$ if all your vectors are on a subspace of  $\mathbb{R}^4$ of dimension $3$ (like the one defined by $B = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}$ )but they'd still have different properties. For example, $2$ orthogonal vectors of  $\mathbb{R}^3$ have only $2$ normalized vectors that are orthogonal to both of them. Whereas in  $\mathbb{R}^4$, there are infinite.
A: first of all dimension means the cardinality of the basis.
Second, if you take out elements from your basis, and they are l.i. vectors, you will span a subspace of your original space. There are elementary theorems about.
Your second question was answered by Chilango. Pay attention to the word "copy". You can make this notion more precisely.
I suggest you to read "Linear Algebra done right", by Sheldon Axler. Probably it will be a good partner.
Greetings.
