Structure of the set of Linear Maps under Functional Composition

Question

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!

Answer 1

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.)

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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.

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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 $C \to D$.