Structure of the set of Linear Maps under Functional Composition So I understand we can clearly form a vector space structure out of the set of linear maps with function addition, but I'm curious about what we can say about this same set with functional composition. I realize we'll have to adjust our set to essentially be two sets, the set of linear maps from V to W and then the set from W to U where V, W, and U are vector spaces. I know we don't get a vector space out of this sort of set up since composition is not commutative. So I'm just wondering what structure we have here or if we can even have a structure with two somewhat different sets. (I'm only an undergrad with background with linear algebra and intro abstract algebra to help responses level of technicality) Thanks!
 A: Wuestenfux in the comment is right, generally one considers the endomorphisms of a vectorspace. Let $V$ be a vectorspace, then the set of endomorphisms is $$\mathrm{End}(V) :=\{f: V \to V | f \text{ linear function}\}.$$
What structure do we have on this set? Well, we have


*

*the identity function $1_V$

*and function composition.


Note that composing with $1_V$ is a no-op! Also function composition is associative. Hence, we have got a monoid on $\mathrm{End}(V)$.
Can we have more? Perhaps a group? Not yet, since we cannot invert wrt. function composition. So let us consider the subset of automorphisms $$\mathrm{Aut}(V) := \{f: V \to V | f \text{ linear bijective function}\}.$$
There we also have inverses $f^{-1}$ for every $f$. (Recall that inverses of linear functions are again linear.) Indeed, we can even add two functions $f, g \in \mathrm{Aut}(V)$ implying that -- skipping some details -- we have a vectorspace structure there. So the set of all automorphisms on a vectorspace forms a vectorspace itself! And yes, you can repeat that: $\mathrm{Aut}(\mathrm{Aut}(V))$ is a vectorspace again. Have a look at $\mathrm{Aut}(\mathrm{Aut}(\ldots(G)))$ asking whether this process ever stabilizes. (I know, it entails math on a level higher than yours -- even mine. But I thought I link it just for you to see what one can play with.)

In full generality, the automorphisms $\mathrm{Aut}(A)$ of an object $A$ in a category form a group! In case of the category of vector spaces, a linear function is just a morphism. A bijective linear function is an isomorphism. So you immediately see how we can express the sets above solely in categorical tonus.

Also if you insist on using two different sets, say the set $$\{f: V \to W | f \text{ linear function}\} \cup \{f: W \to V | f \text{ linear function}\},$$ then you get problems with function composition as the binary operation. Say you have $f, g$ from the right set, then $f \circ g$ does not make much sense. Hence, you have to make your composition operation only defined on "good" operands, or in other words, make it a partial function. That's what happens in a category. The morphism composition in a category is only defined for composable maps $A \to B$ and $B \to C$ -- not for, say, $A \to B$ and $C \to D$.
