Checking whether the 0 vector is in a space spanned by vectors Suppose we have 4 vectors: $v_1 = (4,-4,3,-3)^t, v_2 = (5,-4,4,-3)^t, v_3 = (1,3,-4,-4)^t, v_4 = (0,10,-9,-5)^t$
In my class about linear algebra i've seen a formula to determine the affine space spanned by those vectors, $D = v_1 + span(v_2-v_1, v_3-v_1, ..., v_k - v_1)$. If i use this formula i get $D = (4,-4,3,-3)^t + span((1,0,1,0)^t, (-3,7,-7,-1)^t)$. I've excluded $v_4 - v_1$ because that's a linear combination of the 2 other vectors. 
My first question is, what happens when you do a vector + a span, can you just include the vector in the span if it is not a linear combination of the other vectors, or can't you do that? This is never really addressed in the lectures so i'm not quite sure.
The second question i have is how would i check if 0 is in the affine space spanned by the vectors to see if it is also a vector space? Apparently 0 is not in the affine space (that's in the solution i have), so it's not a vector space but how would i check this? 
 A: In general, given a vector space $V$, an affine subspace is of the form $v + U$ where $U \subseteq V$ is a sub-vector space and $v \in V$ is some vector. The elements of $v + U$ can be characterized as follows: For $w \in V$ it is true that $w \in v + U$ if and only if $w - v \in U$ (use that $w = v + u$ for some $u \in U$).
This is also what is done in your example and explains, in which case $0$ is an element of the affine space, namely, if and only if $v \in U$ (continuing with my notation).
Of course this is not the case in general. An easy example (where you can see it immediately, in contrast to your exercise) might be given by $$A = \begin{pmatrix}1 \\ 0 \end{pmatrix} + span\begin{pmatrix}0\\1\end{pmatrix} \subseteq k^2.$$ As $\begin{pmatrix}1 \\ 0 \end{pmatrix} \notin span \begin{pmatrix}0 \\ 1 \end{pmatrix}$, because these vectors are linearly independent, it follows that $0 \notin A$. (Try to visualize this)
Regarding your first question: This does not that $v$ is just added to the subspace $U$ (in your notation the span of some vectors). As seen in the small example above, the affine space does not even contain one point of the $U$ (or the span).
This is a general fact.
My favourite way to think about affine subspaces is as fibres of linear maps.
Hopefully you have already seen linear maps and kernels: If $A \colon V \to W$  is a linear map and $K = ker(A) = \{v \in V \mid Av = 0\}$ then, for any $w \in W$,  $A^{-1}(w) = \{v \in V \mid Av=w \}$ is either empty or an affine space. If nonempty and $v \in V$ is a preimage of $w$, i.e. $Av = w$, the fibre $A^{-1}(w)$ is characterized as $v + K$, i.e. every preimage $v'$ of $w$ can be written as sum of $v$ and an element of the kernel of $A$.
You can try and picture this for a linear maps $k^3 \to k$ or $k^3 \to k^2$. Then you will see that the fibres are parallel to the kernel. And in particular pairwise disjoint. Only the kernel itself will be a subspace. The others affine spaces which are not vector spaces. Here you just need to be careful with the empty fibres.
