Dilation of Subset of Vector Space Distributes over Sum/Finite Case

Theorem

Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $A_1, \ldots, A_n \subseteq X$ and $\lambda \in \GF$.

Then:

$\ds \lambda \sum_{j \mathop = 1}^n A_j = \sum_{j = 1}^n \paren {\lambda A_j}$

where:

$\ds \lambda \paren \ldots$ denotes dilation by $\lambda$

$\ds \sum_{j \mathop = 1}^n A_j$ denotes the linear combination of subsets of a vector space.

Proof

Let $x \in X$.

We have:

$\ds x \in \lambda \sum_{j \mathop = 1}^n A_j$

if and only if there exists $x_j \in A_j$ for each $j \in \set {1, 2, \ldots, n}$ such that:

$\ds x = \lambda \sum_{j \mathop = 1}^n x_j$

This is equivalent to:

$\ds x = \sum_{j \mathop = 1}^n \lambda x_j \in \sum_{j = 1}^n \paren {\lambda A_j}$ for some $x_j \in A_j$.

Hence we have:

$\ds x \in \lambda \sum_{j \mathop = 1}^n A_j$

if and only if:

$\ds x \in \sum_{j = 1}^n \paren {\lambda A_j}$

Hence we obtain:

$\ds \lambda \sum_{j \mathop = 1}^n A_j = \sum_{j = 1}^n \paren {\lambda A_j}$

$\blacksquare$