Is it possible that there isn't a linear span which precisely spans a vector space? In the assignment I'm asked to decide whether given:
$S = \Bigg \{ \begin{bmatrix}a &b \\ c &d\end{bmatrix} \in M_2(\mathbb{R}) \; | \; ad = 0 \Bigg \},\mathbb{F} = \mathbb{R}$.
$S$ is a vector space.  I believe that addition and multiplication by scalar are defined point-wise.  I believe that this is a vector space, but the question also asks to construct a linear span of this space (provided it is a space).  And here I'm in doubt: any two matrices with $a$ or $d$ non-zero would create elements not in $S$, but if I leave out either $a$ or $d$, the span won't generate every possible matrix.  Is this an okay situation, or did I do something wrong?
 A: Since we're considering $S \subset \Bbb M_2(\Bbb R)$, it is natural to assume that the operations on $S$ are inherited from $\Bbb M_2(\Bbb R)$. But $S$ as it is, is not a subspace of $\Bbb M_2(\Bbb R)$. We have that $$\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1\end{bmatrix}+\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2\end{bmatrix} = \begin{bmatrix} a_1+a_2 & b_1+b_2 \\ c_1+c_2 & d_1+d_2\end{bmatrix}$$
If $a_1=d_2 = 0$ but $a_2,d_1 \neq 0$, closure under addition fails. But $${\rm span} \ S = \bigcap_{S \subset W \leq V} W$$ is always a vector space (intersection of vector spaces is still one). Now, finding a basis for ${\rm span} \ S$ is another problem.
A: Theorem: every vector space has a basis (and any two bases of the vector space have the same cardinality).
This theorem is the cornerstone of linear algebra, and usually the case of finite-dimensional vector spaces is proven in a linear algebra class. (I should warn that the above isn't true in general unless you assume the axiom of choice.) 
Your specific set is not a vector space: $\left( \begin{array}{ccc}
1 & 0\\
0 & 0 \end{array} \right) \in S$ and $\left( \begin{array}{ccc}
0 & 0\\
0 & 1 \end{array} \right)\in S$, but their sum, $\left( \begin{array}{ccc}
1 & 0\\
0 & 1 \end{array} \right) \not\in S$.
