0
$\begingroup$

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: $Sp(S)=<S>$

$\endgroup$
  • $\begingroup$ Does :$Sp(S)=<S>$ ? $\endgroup$ – Mojtaba Jun 14 '15 at 12:16
  • $\begingroup$ Yes $Sp(S)=<S>$ $\endgroup$ – Arcane1729 Jun 14 '15 at 12:23
2
$\begingroup$

$$S\subset \operatorname{Sp(S)}\subset \operatorname{Sp}(S)+\operatorname{Sp}(T),$$ 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: $$\operatorname{Sp}(S)+\operatorname{Sp}(T) \subset\operatorname{Sp}(S\cup T).$$

$\endgroup$
  • $\begingroup$ "remember the span of a subset is the intersection of all the subspaces which contain that subset" - I dont understand this. And how you go on to deduce the next thing $\endgroup$ – Arcane1729 Jun 14 '15 at 12:41
  • $\begingroup$ What is your definition of the span of a subset? $\endgroup$ – Bernard Jun 14 '15 at 12:44
  • $\begingroup$ All linear combinations of that subset $\endgroup$ – Arcane1729 Jun 14 '15 at 12:45
  • $\begingroup$ 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 $S\cup T$. $\endgroup$ – Bernard Jun 14 '15 at 12:59
  • $\begingroup$ 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? $\endgroup$ – Arcane1729 Jun 14 '15 at 13:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.