Linear map dimension So my understanding is a linear map is a function between two vector spaces that satisfies the linear properties w.r.t scalar mult. and vector addition. It is not a vector space itself, so how can it have a dimension? i.e what is meant by $\dim \mathcal L(U,V)$ ?
 A: $\dim\mathcal L(U,V)$ is not the dimension of one particular linear map, but the dimension of the vector space $\mathcal L(U,V)$ whose elements are all the linear maps from $U$ to $V$. This set becomes a vector space because we can add two linear maps:
$$ (f+g)(u) = f(u)+g(u) $$
and multiply a linear map by a scalar:
$$ (c\cdot f)(u) = c\cdot f(u)$$
(and one the needs to check that these operations satisfy the vector space axioms, but that is not hard. They will.)
A: Hint: You can see a linear transformation as a matrix, do you remember?
Thus, the answer is: What is the dimension of the space of matrices?
A: Think about two linear mappings $f$ and $g$, you can definitely define additional two mappings, $\forall u \in U$:
$$h_1(u)=f(u)+g(u),$$
$$h_2(u)=\lambda f(u),$$
for some scalar $\lambda$. If easy to prove, that $h_1$ and $h_2$ are again linear mappings. So we have just defined addition of two mapping and multiplication by scalar.  Therefore set of all linear mappings $\mathcal L(U,V)$ is in fact vector space. 
A: Suppose $U$ and $V$ are vector spaces over a field $\Bbb K$ (think of $\Bbb R$ if you want). First: $${\cal L}(U,V) = \{ T\colon U \to V \mid T \text{ is linear}  \}$$
So ${\cal L}(U,V)$ is a vector space (with the operations defined pointwise), which elements consist of linear maps. Hence, we can speak of $\dim {\cal L}(U,V)$. 
More exactly, let's define: $T + S, c\cdot T \in {\cal L}(U,V)$ by: $$(T+S)(x) := T(x) +_V S(x), \quad (c \cdot T)(x) = c \cdot_V T(x).$$Then one has to check that $T+S$ and $c\cdot T$ really are linear, that is: $$(T+c\cdot S)(x+_U(\lambda\cdot_U y) ) = (T+c\cdot S)(x) +_V \lambda\cdot_V (T+c\cdot S)(y).$$I'm using $+_U$, $+_V$, etc, just to emphasize that the operations are not necessarily the same in different spaces. Usually one uses only $+$ and $\cdot$ (or even omits $\cdot$ at all) for everything and no confusion arises. The properties (associativity, commutativity, etc) for $+$ and $\cdot$ follows from the ones for $+_V$ and $\cdot_V$.
If one has $\dim U = n< +\infty$, $\dim V = m < +\infty$, one can represent $T:U \to V$ by a $m\times n$ matrix, and this association is actually an isomorphism, hence $$\dim_{\Bbb K} {\cal L}(U,V) = \dim_{\Bbb K} {\rm Mat}(m \times n, \Bbb K) = mn = \dim_{\Bbb K}U \dim_{\Bbb K}V.$$
