Summation of Multiple of Mapping on Finite Set

Theorem
Let $\mathbb A$ be one of the standard number systems $\N,\Z,\Q,\R,\C$.

Let $S$ be a finite set.

Let $f : S \to \mathbb A$ be a mapping.

Let $\lambda \in \mathbb A$.

Let $g = \lambda \cdot f$ be the product of $f$ with $\lambda$.

Then we have the equality of summations on finite sets:
 * $\displaystyle \sum_{s \mathop \in S} g(s) = \lambda \cdot \sum_{s \mathop \in S} f(s)$

Outline of Proof
Using the definition of summation on finite set, we reduce this to Indexed Summation of Multiple of Mapping.

Proof
Let $n$ be the cardinality of $S$.

Let $\sigma : \N_{<n} \to S$ be a bijection, where $\N_{<n}$ is an initial segment of the natural numbers.

By definition of summation on finite set, we have to prove the following equality of indexed summations:
 * $\displaystyle \sum_{i \mathop = 0}^{n-1}g(\sigma(i)) = \lambda \cdot \sum_{i \mathop = 0}^{n-1} f(\sigma(i))$

By Multiple of Mapping Composed with Mapping, $g\circ\sigma = \lambda \cdot (f\circ\sigma)$.

The above equality now follows from Indexed Summation of Multiple of Mapping.

Also see

 * Indexed Summation of Multiple of Mapping
 * General Distributivity Theorem