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I apologize if this sounds stupid but I am struggling to grasp the following concept. I understand that the span of the empty set is the zero vector. However, what does the set only containing the zero vector span? The zero vector as well? Also, are the following two phrases logically equivalent, "The span of the empty set is the zero vector" and "The empty set spans the zero vector". I believe they do but I don't want to make any assumptions since I'm not entirely sure myself. Thank you.

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  • $\begingroup$ SPAN$\{\textbf{0}\} = m\textbf{0} = \textbf{0}$ $\endgroup$ – Mr.Fry Feb 21 '15 at 6:36
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    $\begingroup$ Also see this post: math.stackexchange.com/questions/185255/… $\endgroup$ – rundavidrun Feb 21 '15 at 6:59
  • $\begingroup$ Yes, "the span of X is Y" means exactly "X spans Y". But in you example Y is the subspace consisting of the zero vector, not the zero vector itself. $\endgroup$ – Marc van Leeuwen Feb 21 '15 at 11:18
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This is a good question!

In the linear algebra texts that I have seen, it is usually included in the definition of a subspace $S$ that $S$ has to contain the zero-vector. This condition is included for the one purpose of eliminating the empty set as a subspace. (Does your definition do this as well?)

But the span of a set is always a subspace, and in fact, if $L$ is some set, then the span of $L$ is the smallest subspace that contains $L$. This is why it makes sense for the span of the empty set being equal to $\{0\}$, because $\{0\}$ is the smallest subspace containing the empty set (which is not itself a subspace).

Also, if the set $L$ is itself a subspace, then the span of $L$ is equal to $L$ itself - the smallest subspace containing $L$ is $L$. In particular, $\{0\}$ is a subspace, so the span of $\{0\}$ is $\{0\}$.

Regarding your two statements, I say yes, they state the exact same thing.

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  • $\begingroup$ They only state the same thing if we reject the existence of the empty vector space-which in all fairness, most mathematicians do. They accept {0} as the smallest sybspace that can exist. However,there really is nothing logically preventing us from accepting it-but that's when we run into the problems I outlined below. $\endgroup$ – Mathemagician1234 Feb 21 '15 at 9:08
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    $\begingroup$ The defining axioms of a vector space include the existence of a zero-vector (additive identity for vector addition). Thus you can't have an empty vector space. [I don't know if one can deduce the existence of a zero-vector from the other axioms given a nonempty space; but if not, giving up that axiom could be problematic.] If you did have an empty vector space, I wonder what its dimension would be. $\endgroup$ – paw88789 Feb 21 '15 at 10:32
  • $\begingroup$ I would assume it would be 0 by definition,but stranger things have happened in set theory. $\endgroup$ – Mathemagician1234 Feb 21 '15 at 19:19
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In general, adding to a set $S$ of vectors a vector $v$ that is already in the span of $S$ does not change the span; in other words, in this case $S\cup\{v\}$ spans the same subspace as $S$ does.

This applies in particular to $S=\emptyset$ and $v=\vec0$; you already indicated that the span of the empty set contains $\vec0$, and so $\emptyset\cup\{\vec0\}=\{\vec0\}$ spans the same ($0$-dimensional) subspace as the empty set does.

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There are no stupid questions, especially in mathematics by beginning students. (Contrary to what a prominent researcher will tell you in his lectures to make you feel stupid so no other student in the class dares to ask any questions no matter how confused they are-that way,he can run out of the class as fast as possible to get to his real work........lol)

All kidding aside, your question is actually a very good one because it shows you're paying attention. By definition, a subset L of a finite dimensional vector space V spans a subspace S of V iff for every vector $u\in S$, there exist scalars $a_1,a_2,....a_n$ such that with vectors $v_1,v_2,....,v_n\in L$
$$\sum_{i=1}^n {a_i}{v_i}= u $$

There's a reason a lot of linear algebra texts, even the rigorous ones, shy away from this question since it involves some sticky set theory questions. The 2 statements in my humble opinion, are not usually logically equivalent-although why and when they are is a bit subtle.

Consider the empty set. The empty set is a trivial subspace-that is, it satisfies the properties of being a subspace by default. Think about it-if it has no elements, how can it not satisfy the axioms of a vector space by any of it's element? Because a linear combination with arbitrary scalars of no vectors yields zero vectors,the result of such a sum is the zero scalar. Therefore the components of any vector spanned by the empty set are 0 and the only vector this is true of is 0. (Remember, the result of a sum of vectors,even no vectors,must itself be a vector or we don't have a vector space!) Therefore, the empty set spans {0}.

(You might be asking, well, can't any vector space have all zero scalars and this will generate the same result since 0u=0 for any vector u? Actually,no, since {0} is not a field for reasons that are too difficult to discuss here-so this wouldn't be a vector space. ) In the case of the zero vector spanning the empty set, this is trickier. Think about it-if the zero vector spanned $\emptyset$, then there's a linear combination of the zero vector for every vector in $\emptyset$. But by a result in linear algebra,
$$\sum_{i=1}^n {a_i}{0_i}= 0 $$

Where 0 is the 0 scalar. So unless v is a field where the scalars and vectors are interchangable, such as the vector spaces of the real or complex numbers, then the zero vector cannot span 0 since the result of the sum is not the zero vector, but the zero scalar! So the only time the 2 statements can really be equivalent is when the vector space is a field and the vectors and scalars are interchangeable!

I hope you didn't get more confused with my explanation. I hope it's correct-believe me,the other posters will skewer me if it's not. But that's my understanding of it.

Update: Marc van Leeuwen made the following very reasonable criticisms: There is a lot wrong here. "The empty set is a trivial subspace", no it is not a subspace (which should contain at least the zero vector). "Because a linear combination with arbitrary scalars of no vectors yields zero vectors,the result of such a sum is the zero scalar"; mixing vector and scalar zero there. Also working with the empty set, those "arbitrary scalars" have to be 0 in number, that is, absent. "the components of any vector spanned by the empty set" assumes wrongly that vectors must have components. Stopping here because of the length limit in comments...
These problems are precisely why most mathematicians take {0} as the smallest "legal" vector subspace as the intersection of all subspaces in V. Which of course,is a perfectly fine and understandable assumption. But some mathematicians do not because strictly speaking,there's no logical reason to do this. Trivially,empty vectors have empty components and hence the discussion above.Again,though,it's problematic-which is why most mathematicians make this assumption. You can see though that the author of your text or your lecturer-whoever it was-did NOT make this assumption. paw88789 made the additional-and I think much more important-criticism that the axioms of a vector space assume it's a nonempty set, so that technically rules out the empty set as a vector space. I think paw's hit the nail on the head why mathematicians assume a vector space is nonempty-it's precisely to avoid quagmires like this. I think you should bring this up with your teacher and see what he/she thinks.

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  • $\begingroup$ There is a lot wrong here. "The empty set is a trivial subspace", no it is not a subspace (which should contain at least the zero vector). "Because a linear combination with arbitrary scalars of no vectors yields zero vectors,the result of such a sum is the zero scalar"; mixing vector and scalar zero there. Also working with the empty set, those "arbitrary scalars" have to be $0$ in number, that is, absent. "the components of any vector spanned by the empty set" assumes wrongly that vectors must have components. Stopping here because of the length limit in comments... $\endgroup$ – Marc van Leeuwen Feb 21 '15 at 11:31
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    $\begingroup$ These problems are precisely why most mathematicians take {0} as the smallest "legal" vector subspace. Which of course,is a perfectly fine and understandable assumption. But some mathematicians do not because strictly speaking,there's no logical reason to do this. Trivially,empty vectors have empty components and hence the discussion above.Again,though,it's problematic-which is why most mathematicians make this assumption. You can see though that the author of the OP's text-whoever it was-did NOT make this assumption. $\endgroup$ – Mathemagician1234 Feb 21 '15 at 19:18

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