V,W,U are vector spaces over F. Given f and g are linear transformations s.t. f: V -> W and g: W -> U. Show that g o f : V -> U is linear. So this is where i'm at with it... 
$$ \text{So given V,W,U are vector fields over a common field F}$$
$$ \ \ f:V \to W \ \\g:W\to U\\ \text{Are both linear transformations}\\    $$
$$
 \left\{ v_{1},v_{2},...,v_{n} \right\} \text{is an ordered basis of V}
$$
$$
\left\{ w_{1},w_{2},...,w_{m} \right\} \text{is an ordered basis of W}
$$
$$
\left\{ u_{1},u_{2},...,u_{k} \right\} \text{is an ordered basis of U}
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$$$$
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\forall v\in V \ \text{&}\  a,b,c\in F \
$$
$$
 \\ f(v) = f(a_{1}v_{1} + a_{2}v_{2} + \ ...\ + a_{n}v_{n})\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a_{1}f(v_{1}) + a_{2}f(v_{2}) + \ ...\ + a_{n}f(v_{n}) \in W \\ \ \ \ \ \ =b_{1}w_{1} + b_{2}w_{2} + \ ...\ + b_{m}w_{m}
$$
$$ $$
$$ 
g\left( f\left( v \right) \right)=g\left(a_{1}f(v_{1}) + a_{2}f(v_{2}) + \ ...\ + a_{n}f(v_{n})  \right)\\ \ = g\left( b_{1}w_{1} + b_{2}w_{2} + \ ...\ + b_{m}w_{m}  \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   =b_{1}g\left( w_{1} \right) + b_{2}g\left( w_{2} \right) + \ ...\ + b_{m}g\left( w_{m} \right) \in U \\= c_{1}u_{1} + c_{2}u_{2} + \ ...\ + c_{k}u_{k} \ \ \ \ \ \ \ $$
Is this sufficient enough to show that the composition of f and g is also a linear transformation?
 A: No, I'm not a fan of this proof, for a number of reasons.
First of all, there doesn't appear to be an assumption that $V, W,$ or $U$ are finite-dimensional, and hence it's not reasonable to assume the existence of (finite) bases. This result does hold for arbitrary vector spaces, so leaning on bases should be unnecessary.
Secondly, and more problematically, you appear to simply be showing that $g(f(v)) \in U$, choosing to express these in terms of bases. The fact that $g(f(v)) \in U$ is essentially part of the assumptions; if $f : V \to W$ and $g : W \to U$, even if $f$ and $g$ are non-linear, then as part of the definition of $g \circ f$, $g(f(v)) \in U$. To answer your comment, you didn't even use the linearity of $f$ or $g$, you just reiterated the fact that they are maps from $V$ to $W$ and $W$ to $U$ respectively.
Linearity is far stronger than this. You need to show additivity and scalar homogeneity. That is, you need to show that, for any $v_1, v_2 \in V$ (not necessarily elements of a basis), and any scalar $a$,
\begin{align*}
g(f(v_1 + v_2)) &= g(f(v_1)) + g(f(v_2)) \\
g(f(av_1)) &= ag(f(v_1)).
\end{align*}
Don't use bases; just use the fact that $f$ is linear, then use the fact that $g$ is linear. The full proof for each additivity and scalar homogeneity should be no longer than $3$ lines each.
A: If you know a transformation which can be represented by a matrix is linear (and vice-versa), you could just multiply the matrices.
At least, this would handle the finite dimensional case.
