I hope to add more after the question is clarified a little, but here is a start:
If you have a vector space $V$, there will often be lots of subspaces.
For instance, if $v\in V$ is any nonzero element, then $\langle v \rangle=\{\alpha v\mid f \in \mathbb{F}\}$ will be a subspace of $V$, since it's clearly closed under addition and scalar multiplication. It's a one dimensional subspace, so any nonzero vector will generate this little space. If your field is of characteristic $0$, then it will also contain $2v,3v,4v,\dots nv$ because $2,3,\dots n$ are elements of the field $\mathbb{F}$. However there are many more scalars than the integers...
If you pick $w\notin \langle v\rangle$, then $\langle w \rangle$ will produce another subspace of its own, but it won't share any elements with $\langle v \rangle$ except for $0$.
If you take the $v$ and $w$ we have chosen, you can also find another subspace $\langle v,w \rangle=\{\alpha v +\beta w\mid \alpha,\beta\in \mathbb{F}\}$, which is a two dimensional subspace.
It also may be possible to find another $z\in V$ such that $\langle v,z\rangle$, but it is not equal to $\langle v,w\rangle$.
Hopefully you can see how this works for even larger collections of vectors.
If this topic is very new to you, I would highly recommend getting your head around the concept of linear independence as early as possible.
Yes, it is a vector space over F.
A vector space over F is a set V (in your case $m\times n$ matrices with real entries) with operations $+:V\times V \to V$ and $\cdot:F\times V \to V$ satisfying certain axioms, as found for example on the wikipedia vector space page.
That these axioms are satisfied is easily verified. The first four amount to V being an abelian group, and that last four must be satisfied by F since it is subfield of R.
It also follows easily from these observations that if V is a vector space over a field K, then it is a vector space over any subfield F of K (i.e., the example above generalizes).
It is important to note, though, that V is NOT a finite dimensional vector space over F. The dimensional isn't even countable, in fact.
Best Answer
You are right that the set $\{0\}$ cannot be made into a field, because a field must have distinct $0$ and $1$. So, indeed, there is no such thing as a vector space over $\{0\}$; that doesn't mean anything.
However, that is different from $\{0\}$ being a vector space over some field $F$. That is possible and in fact, for every field $F$, $\{0_F\}$ is a zero-dimensional vector space over $F$. Addition and scalar multiplication are defined in the only way they can: everything always results in $0_F$.