14

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.

Martin Sleziak
  • 53,687
Matt Brenneman
  • 1,476
  • 6
    The sum of no vectors is zero. –  Aug 22 '12 at 00:50
  • 2
    In addition to other useful and explanatory answers, observe that Nering says "we agree...". That is, to my interpretation, he expresses disinterest in whether or not this could be proven, or should be a definition, or whatever, ... probably because he sees that it doesn't much matter, and I agree with this. So he is asserting neither, although, as in the answers, your questions can be reasonably addressed, also. In particular, again, in effect he asserts that there is no reason to care much about it, and I agree. – paul garrett Aug 21 '14 at 18:14
  • At the risk of pushing up an already answered question, I'd like to add that since linear span of a set of vectors $S$ is always a vector space and since every vector space contains the $\mathbf{0}$ vector, the linear span of $\mathbf{\phi}$ has to contain $\mathbf{0}$. To prove that it contains only $\mathbf{0}$, it is enough to show it can't contain any other element since those elements aren't present in the set $S$. – Aritra Das Feb 05 '17 at 10:04

5 Answers5

25

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 $\{ 0 \}$.)

Qiaochu Yuan
  • 419,620
  • 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 . We also say that A spans ." – Matt Brenneman Aug 22 '12 at 00:55
  • 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 00:59
  • 1
    @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 01:00
  • 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 01:08
  • 7
    @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 01:10
  • Yes it's a shame as definitions are everything. Thank you very much! – Matt Brenneman Aug 22 '12 at 01:12
  • @Qiaochu Come to think of it, the sum of a set of cardinality 1 also needs a definition, given that addition is a binary operation. I guess we can handle both by saying "write as many +0 as you need until the sum makes sense". –  Aug 22 '12 at 11:45
  • 2
    @LVK: yes, or you can recognize that addition isn't a binary operation; it's a family of $n$-ary operations for every $n \ge 0$, and it just so happens that they can all be defined in terms of the additive identity, the identity operation, and the $n = 2$ operation because of associativity... – Qiaochu Yuan Aug 22 '12 at 16:56
  • Could you please point me a book that has this definition you just gave? – PPP Dec 02 '14 at 01:04
  • @Qiaochu Yuan Could you please refer to a book where I know a little more about the definition of addition you gave. – Lord Shadow Nov 02 '20 at 14:18
4

The span of a set D is the smallest subspace containing the elements of D. Now, every subspace contains 0. Thus if D is a null set the span of D can only be the subspace containing 0.

C. Ravishankar
  • 41
1

It is because Dim({0}) = 0.

Dimension of a vector space refers to the minimum number of basis vector that spans this vector space.

Since Dim({0}) is defined as 0, from the definition of dimension we conclude {0} can be spanned by 0 basis vectors; that is, we must define the span of the empty set as {0} for our definition of dimension to work.

"In the context of vector spaces, the span of an empty set is defined to be the vector space consisting of just the zero vector. This definition is sometimes needed for technical reasons to simplify exposition in certain proofs."

C18
  • 71
0

Looking at the problem from a programatic perspective, if ever I'd like to represent (or generate maybe?) the span of a set of vectors using the linear combinations of vectors definition. I'd always start with a sum variable equating to zero and then iteratively keep on adding to the sum.

If the set would be empty, the script would return the zero vector.

So, Span of an empty set is the zero vector, makes some sense.

swe
  • 21
-1

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.**
Narinder Kumar Dhar
  • 1