📂Linear Algebra

The Creation of Unions is Equal to the Sum of Creations

Theorem¹

Let $S_{1}, S_{2}$ be a subset of the vector space $V$. Then, the following holds.

$$ \span(S_{1} \cup S_{2}) = \span(S_{1}) + \span(S_{2}) $$

Here, $\span$ means generation, and $+$ means the sum of sets.

Proof

• $\span(S_{1} \cup S_{2}) \subset \span(S_{1}) + \span(S_{2})$

  Let $v \in \span(S_{1} \cup S_{2})$. Then, $v$ can be expressed as follows: $$ v = \sum\limits_{i=1}^{n}a_{i}x_{i} + \sum\limits_{j=1}^{m}b_{j}y_{j},\quad x_{i}\in S_{1},\ y_{j} \in S_{2} $$ The first sum belongs to $\span(S_{1})$, and the second sum belongs to $\span(S_{2})$. Therefore, $v \in \span(S_{1}) + \span(S_{2})$.

• $\span(S_{1} \cup S_{2}) \supset \span(S_{1}) + \span(S_{2})$

  $$ u = \sum\limits_{i=1}^{n}a_{i}x_{i} \quad \text{and} \quad v = \sum\limits_{j=1}^{m}b_{j}y_{j},\quad x_{i} \in S_{1},\ y_{j} \in S_{2} $$ Regarding such $u \in \span(S_{1}), v \in \span(S_{2})$, let $u + v \in \span(S_{1}) + \span(S_{2})$. Then, since $u + v = \sum\limits_{i=1}^{n}a_{i}x_{i} + \sum\limits_{j=1}^{m}b_{j}y_{j}$ holds, $u + v \in \span(S_{1} \cup S_{2})$.

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