Linear mapping/Determination on basis/Fact/Proof

Since we want ${\displaystyle {}f(v_{i})=w_{i}}$, and since a linear mapping respects all linear combinations, that is,

${\displaystyle {}f{\left(\sum _{i\in I}s_{i}v_{i}\right)}=\sum _{i\in I}s_{i}f{\left(v_{i}\right)}\,}$

holds, and since every vector ${\displaystyle {}v\in V}$ is such a linear combination, there can exist at most one such linear mapping.
We define now a mapping

${\displaystyle f\colon V\longrightarrow W}$

in the following way: we write every vector ${\displaystyle {}v\in V}$ with the given basis as

${\displaystyle {}v=\sum _{i\in I}s_{i}v_{i}\,}$

(where ${\displaystyle {}s_{i}=0}$ for almost all ${\displaystyle {}i\in I}$) and define

${\displaystyle {}f(v):=\sum _{i\in I}s_{i}w_{i}\,.}$

Since the representation of ${\displaystyle {}v}$ as such a linear combination is unique, this mapping is well-defined. Also, ${\displaystyle {}f(v_{i})=w_{i}}$ is clear.
Linearity. For two vectors ${\displaystyle {}u=\sum _{i\in I}s_{i}v_{i}}$ and ${\displaystyle {}v=\sum _{i\in I}t_{i}v_{i}}$, we have

${\displaystyle {}{\begin{aligned}f{\left(u+v\right)}&=f{\left({\left(\sum _{i\in I}s_{i}v_{i}\right)}+{\left(\sum _{i\in I}t_{i}v_{i}\right)}\right)}\\&=f{\left(\sum _{i\in I}{\left(s_{i}+t_{i}\right)}v_{i}\right)}\\&=\sum _{i\in I}(s_{i}+t_{i})f{\left(v_{i}\right)}\\&=\sum _{i\in I}s_{i}f{\left(v_{i}\right)}+\sum _{i\in I}t_{i}f(v_{i})\\&=f{\left(\sum _{i\in I}s_{i}v_{i}\right)}+f{\left(\sum _{i\in I}t_{i}v_{i}\right)}\\&=f(u)+f(v).\end{aligned}}}$

The compatibility with scalar multiplication is shown in a similar way, see exercise.