A question on linear maps itself Here is just a sample problem.
Suppose that $V$ is finite dimensional and $S, T \in \mathcal{L}(V)$. Prove that $ST = I$ if and only if $TS = I$
Proof : Suppose that $ST = I$. The identity map $I$ is invertible, so by problem 3.22 both $S$ and $T$ are invertible. Multiply $ST = I$ on the right by $T^{−1}$ to get $S = T^{−1}$. We then have $TT^{−1} = TS = I$. Of course the implication $TS = I$ implies $ST = I$ follows by reversing
the roles of $S$ and $T$.
My question : As of my understanding of linear maps, they are functions that map vectors from one vector space to another and function composition may not be commutative, i.e $f \circ g \neq g \circ f$. I am confused on to why they can treat the function as some sort of matrix, multiplying to the left or right sides of functions.
Am I misunderstanding the concept of linear maps or am I missing alternate definitions ?
Thank you for your help, very much appreciated.

Maybe not a good explanation but since $V$ is a vector space over $\mathbb K$ of finite dimension, say , $n$ , we can always view $L(V)$ as matrix algebra $M_n(\mathbb K)$.
The isomorphism is constructed by calculating $f(v_j)$ where $\{v_1,...,v_n\}$ is the basis of $V$, and decomposing them into $f(v_j)=a_{1j}v_1+...+a_{nj}v_n$. Then $(a_{ij})_{i,j}$ is the matrix form of $f$.
The only place where finite dimensional condition is used is "The identity map I is invertible, so by problem 3.22 both S and T are invertible" and the rest of the proof has nothing to do with it. And, whether $f\circ g=g\circ f$ or not, multiplying on both sides should preserve equality (associativity may be needed).
