How to find the basis of a quotient space First of all, I apologize for my level of maths; it's very basic.
I need to find the basis of the quotient space $\Bbb{R}^4/U$
with $U = \{(x_1, x_2, x_3, x_4) \in \Bbb{R}^4 : x_1 + 2x_2 = 0\text{ and } x_3 + 3x_4 = 0\}$
I understand a "basis" to mean a linearly independent (I) spanning set (II) of vectors. 
(I) No vector in the set can be written as a combination of other vectors in the set.
(II) All possible combinations of the vectors in the set will give us our Quotient space (III)
(III) A Quotient space - this isn't a clear concept for me. From the definition here I don't understand what is meant by "collapsing".
Does the fact that I am working with a quotient space change the nature of this problem, or can I just go about trying to find the basis as if i was working with any old vector space? If so, could someone give me a hint as to how I can do this? 
I think it involves solving the equations $x_1 + 2x_2 = 0, x_3 + 3x_4 = 0$. Is this the correct direction? 
Many thanks in advance.
 A: To find a basis for a quotient space, you should start with a basis for the space you are quotienting by (i.e. $U$). Then take a basis (or spanning set) for the whole vector space (i.e. $V=\mathbb{R}^4$) and see what vectors stay independent when added to your original basis for $U$.
In your case, it's pretty easy to see that $\{ (2,-1,0,0), (0,0,3,-1) \}$ form a basis for $U$ (essentially $U$ is a null space - you could use Gaussian elimination to find such a basis). We now need $\dim(\mathbb{R}^4)-\dim(U)= 4-2=2$ more vectors to complete this to a basis for $\mathbb{R}^4$. It should be fairly obvious that $(1,0,0,0)$ and $(0,0,1,0)$ are independent of our basis for $U$ (to verify you could always chuck all of these vectors into a matrix, row reduce, and see that you get the identity matrix). 
Ok. We now have a basis $\{ (2,-1,0,0), (0,0,3,-1), (1,0,0,0), (0,0,1,0) \}$ for $\mathbb{R}^4$ where the first two vectors form a basis for $U$. The second two vectors are independent of the first two. These vectors give representatives for cosets which form a basis for $\mathbb{R}^4/U$. That is, $\{ (1,0,0,0)+U, (0,0,1,0)+U \}$ is a basis for $\mathbb{R}^4/U$.
Here's more or less an algorithm: Suppose $U$ is a subspace of $V$. Let $A=[a_1\;a_2\;\cdots\;a_m]$ be a matrix whose columns span $U$. Let $B = [b_1\;b_2\;\cdots\;b_n]$ be a matrix whose columns span $V$. Row reduce the matrix $[A\;B] = [a_1\;\cdots\;a_m\;b_1\;\cdots\;b_n]$ to locate pivot columns. Say among the $a_i$'s, $c_1,\dots,c_k$ are pivots and among the $b_j$'s, $d_1,\dots,d_\ell$ are pivots. Then $\{c_1,\dots,c_k\}$ form a basis for $U$ and $\{d_1+U,\dots,d_\ell+U\}$ form a basis for $V/U$.
