Are all matrices linear operators? Given $A \in \mathbb{K}^{n\times m}$ a matrix, can we think of $A$ as an operator?
In what context do matrices satisfy the definition of operator?
 A: They do not satisfy the definition of a linear map. Instead, there is a canonical bijection:
$$\text{Vect}(\mathbb{K}^m,\mathbb{K}^n) \to \mathbb{K}^{n\times m}$$
where $\text{Vect}(\mathbb{K}^m,\mathbb{K}^n)$ is the set of linear maps from $\mathbb{K}^m$ to $\mathbb{K}^n$. It is given by mapping a linear map $f : \mathbb{K}^m \to \mathbb{K}^n$ to the matrix $(f(e_1), \dots, f(e_m))$. Its inverse maps a matrix $A$ to the map $f$ with $f(v) = Av$ for all $v\in \mathbb{K}^m$.
So, it doesn't matter, whether we study matrices or linear maps, at least in this case. If you consider other vector spaces without 'standard bases' it gets more complicated.
A: In addition to Stefan Perko's answer..
The square matrices are not linear maps.
They rather form an algebra:
The simplest examples of unital C*-algebras.
Now, there is the Gelfand-Naimark theorem:
It says that they can be interpreted as continuous linear operators over some Hilbert space.
If I'm not mistaken, this turns out to be the canonical bijection described by Stefan Perko.
