Theorem 1 on Linear Mappings

Let $V$ and $W$ be $K$-vector spaces. Let $b_1, \ldots, b_n$ be a basis of $V$ and $w_1, \ldots, w_n$ be arbitrary vectors in $W$. Then there is exactly one linear mapping

f : V \to W

with

f(b_i) = w_i

for all $i = 1, \ldots, n.$

Beweis

We define $f$ by the linear extension of the condition $f(b_i) = w_i$ for $i = 1, \ldots, n$. If $v = \sum_{i=1}^n \lambda_i \cdot b_i \in V$, then we set

f(v) = f \left( \sum_{i=1}^n \lambda_i b_i \right) = \sum_{i=1}^n \lambda_i f(b_i) = \sum_{i=1}^n \lambda_i w_i.

Due to Remark 1 on Linear Mappings, the mapping $f$ is well-defined. As can be easily verified, $f$ is linear. Because $f(b_i) = w_i$ for all $i = 1, \ldots, n$ and due to linearity, $f$ is uniquely determined. $\square$