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I am reading Nering's book on Linear Algebra and in the section on vector spaces he makes the comment, "We also agree that the empty set spans the set consisting of the zero vector alone".

Is Nering defining the span of the empty set to be the set containing the zero vector or is this something you can prove from the definition of span? I sense it is the latter, but the proof seems a bit tricky since you would be saying that {0} = Span of the indexed set of vectors in the empty set. But since the empty set has no vectors, it is not clear to me what its span would be.

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The sum of no vectors is zero. –  Rahul Aug 22 '12 at 0:50
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up vote 8 down vote accepted

Depending on how you define the span, this is either a definition or it follows from the definition of span (and judging by the wording it is probably the former). What's Nering's definition of span?

(One definition of span is the following: the span of a collection of vectors is the intersection of all subspaces containing them. The span of no vectors is therefore the intersection of all subspaces, which is $\{ 0 \}$.)

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This is the definition Nering gives for the span of a subset A of the vector space V: "For any subset A of V the set of all linear combinations of vectors in A is called the set spanned by A, and we denote it by <A>. We also say that A spans <A>." –  Matt Brenneman Aug 22 '12 at 0:55
BTW, I like your definition. Technically that would solve my problem, but I still wouldn't understand my problem ;>) –  Matt Brenneman Aug 22 '12 at 0:59
@Matt: that definition is ambiguous unless Nering has also defined what a linear combination of no vectors is (as Rahul says in the comments, the correct definition is that it is $0$). I am guessing that Nering has not defined a linear combination of no vectors, and so this is an additional definition. –  Qiaochu Yuan Aug 22 '12 at 1:00
As far as I can tell, he makes no such definition. He defines a linear combination as a finite sum of indexed vectors each multiplied by an indexed field element, and that's it. You can see why this is a problem for me, b/c there is no such indexed set for any empty set, so the definition of "linear combination" seems hard to apply to a set which can not be represented as an indexed set. The only other mention he makes of the empty set is to say that the empty set is linearly independent (which I reasoned as being true b/c it is vacuously true). –  Matt Brenneman Aug 22 '12 at 1:08
@Matt: yes, the problem is that most authors do not define the empty sum. The empty sum should be $0$. –  Qiaochu Yuan Aug 22 '12 at 1:10
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linear span of an empty set i.e L(0) is taken as the set (O),this is confusing because L(0) is the set of all linear combinations of the elements of 0 but to make a linear combination we need to have at least one vector of the set and empty set contains no vectors in it. Thus it it should have been 0 and not (O).

On the other hand, if possible, let L(0) be a set other than (O). Then then it* either contains at least one non zero element (i.e a vector of V) which is a linear combination of the elements of 0.This leads us to a contradiction that 0 is not empty.Hence the possibility of L(0) to be other than (O) is ruled out.

  • or is it self 0.Now if this possibility can be ruled out then the proof becomes complete. reader may please comment.**
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