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.
 A: $ST = I$ is an abbreviation for $S \circ T = I$. In this case, by multiplying on the right by $T^{-1}$, they mean
$$(\star) \hspace{0.7cm} (S \circ T) \circ T^{-1} = I \circ T^{-1}. $$
Notice that $S, T \in \mathcal{L}(V)$, so both $S \circ T$ and $T \circ S$ are defined. Moreover, the condition that $S \circ T = I$ (or $T \circ S = I$) implies that both $S$ and $T$ are invertible, so their inverses $S^{-1}$ and $T^{-1}$ are also linear maps from $V$ to $V$. This makes the composition in $(\star)$ legitimate.
This result is saying that for linear maps from a finite-dimensional vector space $V$ to itself, it is enough to have either $S \circ T = I$ or $T \circ S = I$ to deduce the other.

On the other hand, there is a relation between composition of linear maps and matrix multiplication. Suppose that $U, V, W$ are finite-dimensional vector spaces and choose a basis $\mathcal{B}$ for $U$, a basis $\mathcal{E}$ for $V$, and a basis $\mathcal{D}$ for $W$. If $T: U \rightarrow V$ and $S: V \rightarrow W$ are linear maps, then $S \circ T$ is a linear map from $U$ to $W$, and we can calculate its matrix with respect to the bases of $U$ and $W$ via matrix multiplication:
$$ M_{\mathcal{B}}^{\mathcal{D}}( S \circ T) = M_{\mathcal{E}}^{\mathcal{D}}( S ) M_{\mathcal{B}}^{\mathcal{E}}( T ), $$
so the matrix of $S \circ T$ is the product of the matrices of $S$ and $T$.
A: I'll point out some basic facts about linear maps that might be helpful.
Consider some real matrix $A$. It's easy to verify that the function $x \mapsto Ax$ is a linear map. On the other hand it's possible to show that given any linear map $f: \mathbb{R}^n \to \mathbb{R}^m$, there is some matrix $A$ such that $f(x) = Ax$. In other words, matrices and linear maps are two aspects of the same thing.
About the matrix product: Consider two linear maps $f$ and $g$. Let $A$ and $B$ the corresponding matrices. What's the matrix of $f \circ g$? Well it's the product matrix $A B$. Why? Well ever wonder why matrix product is defined in such a weird way? The whole point in defining it that way is to make $AB$ correspond to $f \circ g$.
A: 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).
