Let $S$ and $T$ be two subsets of a vector space $E$. I want to prove

$span(S) + span(T) = span(S \cup T)$.

Here is my approach: let $span(S) = span(v_1, \dots, v_m)$ and $span(T) = span(u_1, \dots, u_n)$.

  • $span(S) + span(T) \subseteq span(S \cup T)$. Let $w \in span(S) + span(T)$. Then $w = x+y$, where $x \in span(S)$ and $y \in span(T)$. Since $x \in span(S)$, $x = \alpha_1 v_1 + \dots + \alpha_m v_m$ for some $\alpha_i \in \mathbb{R}$, $i=1, \dots, m$. Similarly, $y = \beta_1 u_1 + \dots + \beta_n u_n$ for some $\beta_i \in \mathbb{R}$, $i = 1, \dots, n$. Therefore, since $w = \alpha_1 v_1 + \dots + \alpha_m v_m + \beta_1 u_1 + \dots + \beta_n u_n$, $w \in span(v_1, \dots, v_m, u_1, \dots, u_n$) i.e. (not sure if the following step is right) $w \in span(S\cup T)$.
  • $span(S \cup T)\subseteq span(S) + span(T)$. Let $w \in span(S \cup T)$, then $w = \alpha_1 v_1 + \dots + \alpha_m v_m + \beta_1 u_1 + \dots + \beta_m u_m = x+y$, where $x \in span(S)$ and $y \in span(T)$. Therefore, $w \in span(S) + span(T)$.

Im not sure if those steps are right, mainly the ones involving the fact that a vector belongs to the union of subsets. Is this correct or is there another way, more elegant maybe, to prove this? Thank you!


1 Answer 1


That is not correct. You are assuming that $S$ and $T$ are finite, but that is not part of the assumptions of the problem.

Since $S\subset S\cup T$, $\operatorname{span}(S)\subset\operatorname{span}(S\cup T)$. For the same reason, $\operatorname{span}(T)\subset\operatorname{span}(S\cup T)$, and therefore$$\operatorname{span}(S)+\operatorname{span}(T)\subset\operatorname{span}(S\cup T).$$

On the other hand, if $v\in\operatorname{span}(S\cup T)$, then there are vectors $v_1,\ldots,v_k\in S$, there are vectors $v_{k+1},\ldots,v_l\in T$ and there are scalars $\lambda_1,\ldots,\lambda_l$ such that$$v=\overbrace{\lambda_1v_1+\cdots+\lambda_lv_k}^{\phantom{\operatorname{span}(S)}\in\operatorname{span}(S)}+\overbrace{\lambda_{k+1}v_{k+1}+\cdots+\lambda_lv_l}^{\phantom{\operatorname{span}(T)}\in\operatorname{span}(T)},$$and therefore $v\in\operatorname{span}(S)+\operatorname{span}(T)$.

  • $\begingroup$ Thank you very much! You are right I assumed things that were not said, thanks again! $\endgroup$
    – Uri Toti
    Feb 12, 2021 at 9:22
  • $\begingroup$ By the way, just to clarify myself. $span(S) + span(T) \subset span(S\cup T)$ because $span(S\cup T)$ is a subspace and therefore closed under addition, is this reasoning correct? $\endgroup$
    – Uri Toti
    Feb 12, 2021 at 9:27
  • $\begingroup$ Yes, that is correct. $\endgroup$ Feb 12, 2021 at 9:28
  • $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ Feb 12, 2021 at 9:33
  • $\begingroup$ StackExchange doesn't allow me until 3 minutes have passed, I have already tried. $\endgroup$
    – Uri Toti
    Feb 12, 2021 at 9:34

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