Defining composition of multiples of functions. There's this following theorem in my book whose proof is left as an exercise.
Theorem: Let $V(F)$ be a vector space and let $U_1,U_2\in\mathcal{L}(V)$. Then, $\forall a\in F, a(U_1U_2)=(aU_1)U_2=U_1(aU_2)$.
Here, $\mathcal{L}(V)$ denotes the vector space of all linear transformations from $V$ to $V$ and $AB$ (also denoted by $A\circ B)$ denotes the composition of two linear transformations $A$ and $B$.
Now, I know that there is a bijection between matrices and linear transformations and as such, I can view the theorem in an alternative way where the theorem is simply equivalent to the product of two matrices $U_1,U_2$ and scalar $a$, which makes the theorem look trivial.
But, I'm having a bit of a problem proving this while considering $U_1,U_2$ as linear transformations.
I have established equality between LHS and MHS but I can't establish the equality with the RHS.

This made me ponder over something related to this that might point out the error in my thinking.
Consider functions $f,g\colon A\mapsto A$ and consider real scalars $a,b$. Then, how exactly do we define $(af\circ bg)(x)$ ?
Do we define it as $a\cdot f(b\cdot g(x))$ or do we define it any other way?
I've been using the above definition in my work. Maybe if this definition is incorrect, that might be the reason for my error? Can you guys help me out? Thanks in advance! :)
 A: If $U_1$ is a linear transformation then for $a \in F$, it's true for any $v \in V$ that $aU_1(v) = U_1(av)$. In particular it's true for $v = U_2(x), x \in V$. That is,
$$aU_1(U_2(x)) = U_1(aU_2(x))$$
for all $x \in V$. That's the remaining piece of the puzzle.
It's also possible to prove this without explicitly using elements of $V$. The vector space axioms, and the fact that field multiplication is commutative, show that for any $a \in F$, 


*

*the function $T_a \colon x \mapsto ax \colon  V \to V$ is a linear transformation, such that

*if $U$ is any linear transformation of $V$, then $T_a \circ U = U \circ T_a$.


(You have to use elements of $V$ to show that.) Using the fact that composition of functions is associative, and the fact that $aU = T_a \circ U$, the result now follows easily:
$$
\begin{align}
a (U_1 U_2) &= T_a \circ (U_1 \circ U_2) \\
&= (T_a \circ U_1) \circ U_2 \\
&= (U_1 \circ T_a) \circ U_2 \\
&= U_1 \circ (T_a \circ U_2) \\
&= U_1 (a U_2) \\
\end{align}
$$
Usually you'd just write $a$ for the linear transformation $x \mapsto ax$, by a slight "abuse of notation", but more explicit/pedantic notation can be clearer when you're first learning.
