Definition talk:Generated Submodule/Linear Span

The definition is equivalent to two others, similar to what is done for Definition:Closed Linear Span. --Lord_Farin 10:05, 4 January 2012 (CST)

Hm, GFP, be sure that you understand that the notion of Linear span can be associated to any set of vectors in $n$-space. If this set is finite, then the sum indeed can be taken over all elements. However, to only define the span of $n$ elements in $n$-space is too narrow. Please modify it or reply. --Lord_Farin 03:11, 30 January 2012 (EST)

I've only come across it with finite vectors, Good now? --GFauxPas 06:29, 30 January 2012 (EST)

Revisiting the issue now that my class addressed infinite sets. Why can't we define the span as the set of all linear combos of the set, i.e., every linear combo of a finite sequence of vectors in the space? --GFauxPas 11:11, 20 April 2012 (EDT)

I am quite sure that that is precisely what is expressed in the definition at the moment... so maybe I misunderstood you. --Lord_Farin 18:52, 20 April 2012 (EDT)

Well you didn't misunderstand me because when I first edited the page I didn't know about infinite-dimensional spaces yet. Anyway, "When dealing with a finite set $A$ of vectors, the linear span can be interpreted as the set of all linear combinations of these vectors.", my question is, can't I delete "When dealing with a finite set $A$ of vectors"? --GFauxPas 19:01, 20 April 2012 (EDT)

Entirely correct, as long as a linear combination is stipulated finite (which it is). --Lord_Farin 19:09, 20 April 2012 (EDT)

Alrighty then. --GFauxPas 19:14, 20 April 2012 (EDT)

Can $A = \O$? I don't see how that would work. It's related to my question here, because if a subspace has to be non-empty then how can we say Linear Span is Linear Subspace if $A$ might be empty? --GFauxPas 15:01, 11 March 2012 (EDT)

Don't forget the empty sum. By default, it equals the identity element of addition. This might deserve explicit statement on the proof page, though. --Lord_Farin 19:01, 11 March 2012 (EDT)