Action of Linear Operator on basis of a vector space. Say I have a vector space $V$ having basis vectors $\vec{e}_i \text{ for } 1\le i\le n$.
Then, the action of a linear operator $f$ on the basis vectors is given by:
$$f(\vec{e}_i) = \sum_{k=1}^n f_{ki}\hat{e}_k$$
I wanted to learn the reason behind this definition.
I read some articles on the internet. Some have stated this as a definition, whereas some stated that since $f(\vec{e}_i)$ is in $V$, it should be written as a linear combination of basis vectors which gives the above result.
Can anyone help me out with this confusion?
The related articles:
http://www.physics.miami.edu/~nearing/mathmethods/operators.pdf Eqn: 7.6
https://ocw.mit.edu/courses/8-05-quantum-physics-ii-fall-2013/04b0570b349e84d74129eef504498472_MIT8_05F13_Chap_03.pdf Eqn 2.26
 A: The theorem you want is that for any set $\{f_{ki} \vert  1 \le i,k \le n \} $of scalars there exists exactly one linear map $f:V \to V$ satisfying your equations. Now you can ask if this theorem can be extended to infinite- dimensional vector spaces and what about modules or free modules or free modules of finite rank over a commutative ring (with identity)?
A: Let your linear transformation be $f:V\to W$ where we are
taking letters $b_1,...,b_n$ for basics of $V$ and $c_1,...,c_m$ for basics on $W$
When you know what a linear transformation is, then
you can assign to basic vector in the domain as your
$$f(b_1)=f_{11}c_1+...+f_{m1}c_m$$
$$f(b_2)=f_{12}c_1+...+f_{m2}c_m$$
$$\cdots$$
$$f(b_n)=f_{1n}c_1+...+f_{mn}c_m,$$
we see how the basis in $V$ is transformed.
These linear combinations exist because the $c_s$ generate $W$.
With this now, it's easy to find how a generic $v\in V$ is transformed,
that is via
$$f(v)=f\left(\sum_sv_sb_s\right)$$
$$=\sum_sv_sf(b_s)$$
$$\quad=\sum_s v_s\left(\sum_t f_{ts}c_t\right)$$
$$=\sum_t \left(\sum_sv_sf_{ts}\right)c_t,$$
place where we can see the components of $f(v)$ in the basis $c_t$ of $W$.
