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

$f:V⟶W$

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

$\begin{array}{cc}f(u+v)& =f(\left(\sum _{i\in I}{s}_i{v}_i\right)+\left(\sum _{i\in I}{t}_i{v}_i\right))\\ & =f\left(\sum _{i\in I}(s_i+t_i)v_i\right)\\ & =\sum _{i\in I}(s_i+t_i)f\left(v_i\right)\\ & =\sum _{i\in I}{s}_{i}f\left(v_i\right)+\sum _{i\in I}{t}_{i}f(v_i)\\ & =f\left(\sum _{i\in I}{s}_i{v}_i\right)+f\left(\sum _{i\in I}{t}_i{v}_i\right)\\ & =f(u)+f(v).\end{array}$

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

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