# Linear Algebra- Sums of Vector Spaces

I dont know how to prove this although intuitively I know that it is true:

Let $V$ be a finite dimensional vector space and $S$ and $T$ be subsets of $V$.

Show that $$Sp(S\cup T) = Sp(S)+Sp(T)$$ I think I have to show both sides are subsets of each other but I'm not sure how. I take it the question means sum of two vector spaces for the RHS

Edit: $Sp(S)=<S>$

• Does :$Sp(S)=<S>$ ? – Mojtaba Jun 14 '15 at 12:16
• Yes $Sp(S)=<S>$ – Arcane1729 Jun 14 '15 at 12:23

$$S\subset \operatorname{Sp(S)}\subset \operatorname{Sp}(S)+\operatorname{Sp}(T),$$ and similarly $\,T\subset\operatorname{Sp}(S)+\operatorname{Sp}(T)$, whence $S\cup T \subset \operatorname{Sp}(S)+\operatorname{Sp}(T)$, and finally (remember the span of a subset is the intersection of all the subspaces which contain that subset): $$\operatorname{Sp}(S\cup T) \subset \operatorname{Sp}(S)+\operatorname{Sp}(T).$$
Conversely, as $\,S\subset S\cup T$, $\,\operatorname{Sp}(S)\subset\operatorname{Sp}(S\cup T)$. The same is true for $T$, so that: $$\operatorname{Sp}(S)+\operatorname{Sp}(T) \subset\operatorname{Sp}(S\cup T).$$
• In that case my parenthetical remark is useless. It's evident that if $S\cup T$ is contained in a subspace, its span is also contained in the subspace. But the characterisation I mentioned is true, and can be the useful point of view in some case. You also can say it is the smallest (for inclusion) subspace that contain $S\cup T$. – Bernard Jun 14 '15 at 12:59
• ahhhhh- because each span on the RHS is a subspace and the sum of two subspaces is a subspace? $S\cup T$ is contained in a subspace- so its span is contained in that subspace? – Arcane1729 Jun 14 '15 at 13:02