Linear mapping/Determination on basis/Fact/Proof

Proof

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

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

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

$f\colon V\longrightarrow W$

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

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

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

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

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

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

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

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