Suppose we want to find a basis for the vector space $\{0\}$.

I know that the answer is that the only basis is the empty set.

Is this answer a definition itself or it is a result of the definitions for linearly independent/dependent sets and Spanning/Generating sets? If it is a result then would you mind mentioning the definitions of bold items which based on them this answer can be deduced.

Useful Links

These are the links that I found useful for answering this question. It needs some elementary background form mathematical logic. You can learn it by spending a few hours on this Wikipedia page.

Link 1, Link 2, Link 3, Link 4, Link 5, Link 6

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    $\begingroup$ Here is a related question: How many nonzero vectors does the zero vector space have? (an element of a basis has to be a nonzero vector by definition) $\endgroup$ – Fabian Aug 31 '16 at 11:53
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    $\begingroup$ @H.R.: Related: math.stackexchange.com/questions/1812653 $\endgroup$ – Watson Aug 31 '16 at 12:02
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    $\begingroup$ Maybe I did not understand your question well: My comment was just intended to indicate that only $\{\}$ can potentially be a basis. Of course you then should show that it is indeed a basis (or you have already proven that any finite dimensional vector space indeed has a basis). $\endgroup$ – Fabian Aug 31 '16 at 12:11
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    $\begingroup$ Vacuous statements are true. You will see this type of argument frequently occuring though pointless to me. $\endgroup$ – IAmNoOne Aug 31 '16 at 13:12
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    $\begingroup$ @H.R. math.stackexchange.com/questions/734418/… $\endgroup$ – IAmNoOne Aug 31 '16 at 13:21

The standard definition of basis in vector spaces is:

$\mathcal B$ is a basis of a space $X$ if:

  • $\mathcal B$ is linearly independent.
  • The span of $\mathcal B$ is $X$.

You can easily show both of these statements are true when $X=\{0\}$ and $\mathcal B= \{\}$. Again, you have to look at the definitions:

  • Is $\{\}$ linearly independent? Well, a set $A$ is linearly independent if, for every nonempty finite subset $\{a_1,a_2\dots, a_n\}$, we have that if $$\alpha_1a_1 + \dots + \alpha_n a_n=0,$$ then $\alpha_i=0$ for all $i$. This condition is satisfied automaticall in the case of an empty set (everything follows from a false statement). This part may be difficult to understand, but since there is no nonempty finite collection of vectors from $\{\}$, any statement you say about nonempty finite collections of vectors from $\{\}$ must be true (because any such statement includes an assumption that a nonempty finite collection exists. It does not, meaning that any such statement is of the type $F\to A$ and is automatically true). This means $\{\}$ is linearly independent.

  • Is the span of $\{\}$ equal to $\{0\}$? Well, the span of a set $A\subseteq X$ is defined as the smallest vector subspace of $X$ that contains $A$. Since all vector subspaces contain $\{\}$, it is clear that $\{0\}$, which is the smallest vector subspace at all, must be the span of $\{\}$.

Alternatively, the span of $A$ is the intersection of all vector subspaces that contain $A$. Again, it should be obvious that this implies that the span of $\{\}$ is $\{0\}$.

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    $\begingroup$ On the linearly independent issue, I would have said that the sum of zero terms is defined as $0$, so that the only subset, $\{\}$ give equation $0 = 0$ and we don't have to deal with ugly false statements in a proof. $\endgroup$ – Lærne Aug 31 '16 at 12:07
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    $\begingroup$ @Lærne Don't get me wrong, both definitions work, but none of them is trivial, so it's a matter of opinion which is best. $\endgroup$ – 5xum Aug 31 '16 at 12:26
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    $\begingroup$ @H.R. Precisely. Since there is no finite collection of vectors from $\{\}$, any statement you say about finite collections of vectors from $\{\}$ must be true! $\endgroup$ – 5xum Aug 31 '16 at 18:32
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    $\begingroup$ @H.R. Because the statement "$A$ is linearly dependend" does not start with "for any nonempty subset of $A$". Thather, the statement starts with "There exists a nontmpty subset of $A$", and such a statement is clearly false. $\endgroup$ – 5xum Sep 1 '16 at 8:55
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    $\begingroup$ @H.R. No, in fact, it should say "finite non-empty". $\endgroup$ – 5xum Sep 1 '16 at 9:29

Definition 1. The span of a set of vectors $\{v_1,\ldots,v_m\}$ is the set of all linear combinations of $\{v_1,\ldots,v_m\}$. In other words, $$\text{span}\{v_1,\ldots,v_m\}=\{a_1v_1+\cdots+a_mv_m,\, a_1,\ldots,a_m\in\mathbb{F}\}.$$

This definition leaves out the case for $\{\}$: there is no vector to begin with! So we need to take care of that. But how do we define the span of $\{\}$? We define it to be $\{\}$? Or some arbitrary space? Here is the rationale for defining $\text{span}\{\}$ to be $\{0\}$:

Proposition. Let $V$ be a vector space. Let $S$ be a finite subset of $V$ that spans $V$. One can obtain a basis of $V$ by deleting elements from $S$.

Only then can we have this proposition working for $V=\{0\}$.

To summarize, when our definition of span is as in Definition 1, we want the following extra definition

  1. The empty set is independent;
  2. The span of the empty set is the zero space $\{0\}$

for the above proposition to be true for $V=\{0\}$. As a consequence of our definition, the empty set is a basis for the zero vector space.

(Notes: My definition of linear independence is:

A set of vectors $\{v_1,\ldots,v_m\}$ is said to be linearly independent if the equation $a_1v_1+\cdots+a_mv_m=0$ always implies $a_1=\cdots=a_m=0$. Otherwise, it is said to be linearly dependent.

And I do not define the "empty sum", so that the case $\{\}$ is left undetermined. )

Definition 2. The span of a set of vectors $\{v_1,\ldots,v_m\}$ is the smallest vector space containing $v_1,\ldots,v_m$.

Under this definition, indeed we do not need to additionally define the span for $\{\}$, as @5xum pointed out.

Definition 1 is more common, since elements of the set $\text{span}\{v_1,\ldots,v_m\}$ are described explicitly. The drawback of Definition 2 is that you don't know what the elements in the span look like, and you need to prove that the span of $\{v_1,\ldots,v_m\}$ indeed consists of linear combinations of $v_1,\ldots,v_m$.

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    $\begingroup$ I object to using the term "need to define". You don't need to define that. There are perfectly valid definitions of "independent" and "span" that directly imply that $\{\}$ is independent and that its span is $\{0\}$. $\endgroup$ – 5xum Aug 31 '16 at 12:32
  • $\begingroup$ Note that your definition of "span" is different from here's. There is no way to deduce the implication and you need to take care for the $\{\}$. $\endgroup$ – Fei Li Aug 31 '16 at 12:35
  • $\begingroup$ So the OP's question ultimately depends on the definition of span. $\endgroup$ – Fei Li Aug 31 '16 at 12:36
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    $\begingroup$ Note that if one first defines linear dependence for a set of $m$ vectors, namely there exists $m$ scalars, not all zero, such that the linear combination $\sum a_iv_i$ is zero, and otherwise to be linearly independent, then the linear independence of $\{\}$ would immediately follow. $\endgroup$ – Fei Li Sep 1 '16 at 11:50
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    $\begingroup$ I can perfectly see the reason for defining the empty sum to be zero. And note two things: (1) most linear algebra textbooks wouldn't bother to interrupt the discussion of linear algebra to talk about the "empty sum"; (2) if one defines linear dependence as in my above comment, and defines span as in Definition 2, then we can deduce that $\{\}$ is linearly independent and spans $\{0\}$, thus is a basis for $\{0\}$, without resorting to empty sum. $\endgroup$ – Fei Li Sep 1 '16 at 11:51

To answer the question of why $\text{Span}\{\}=\{0\}$ is true, I considered the following argument for myself.

I think all the things draw back to the operation of addition.

Addition is a map defined as follows

$$ \begin{align} (.+.): & V \times V \to V \\ & (u,v) \to (u+v) \end{align}$$

with the commutative and associative properties

$$ \begin{align} (u+v) &= (v+u) \\ ((u+v)+w) &= (u+(v+w)) \end{align} $$

so according to this definition, whenever we are talking about addition we should provide two inputs to get one output.

From a programming point of view it is useful to have outputs in case when we have one or no inputs (see the detail of Plus function in wolfram language). It also turns out to be useful in proofs like the ones using induction. So what is the most useful definitions to make for such cases? Experience shows that these are

$$ \begin{align} (u+\text{null}) &= u\\ (\text{null}+u) &= u \\ (\text{null}+\text{null}) &= 0 \end{align} \tag{1}$$

You can also think that when we don't provide enough inputs to the addition function then it assumes $0$ instead of the arguments which are not provided namely the $\text{null}$ arguments.

Now the following definition can be easily interpreted in the special cases when we have a set with one element or no element.

Linear Combination and Span. A linear combination of a set $A=\{v_1,v_2,...,v_m\} \subseteq V$ is a vector $v$ defined by $v=\sum_{j=1}^{m}a_jv_j$. The set of all linear combinations of $A$ is called the span of $A$ denoted by $\text{Span}A$.

Now, if we make the convention that $m=0$ means $A=\{\}$ and when $m=1$ then $A=\{v_1\}$, according to $(1)$, we can interpret the definition as follows

$$ v=\sum_{j=1}^{m}a_jv_j:=s_m, \qquad s_i = \begin{cases} (\text{null}+\text{null}), & i=0 \\ (a_1v_1+\text{null}), & i=1 \\ (a_1v_1+a_2v_2), & i=2 \\ (s_{i-1}+a_{i}v_{i}), & \text{Otherwise} \\ \end{cases} , \qquad 0 \le i \le m $$

So, we can see that $\text{Span}\{\}=\{0\}$. Note that this is a result of our own convention for the addition operation and the definition of the span.

  • $\begingroup$ I would say this interpretation is... unwise. It is simpler to define a linear combination $v = \sum_{j=1}^m a_jv_j$ as $s_m$, where $s_i = \begin{cases}0 & \text{if $i=0$}, \\ s_{i-1}+a_iv_i & \text{otherwise.}\end{cases}$ $\endgroup$ – Rahul Nov 6 '17 at 22:06
  • $\begingroup$ @Rahul: Thanks for the attention. :) You are right. :) You wrote the last equation more briefly and efficiently. :) $\endgroup$ – H. R. Nov 6 '17 at 22:15
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    $\begingroup$ And this way you don't have to redefine vector addition to act on zero or one operands; we still always add exactly two vectors at a time :) $\endgroup$ – Rahul Nov 6 '17 at 22:48
  • $\begingroup$ @Rahul: Indeed, my explanation after Eq.$(1)$ is exactly what you said. :) Maybe I wrote it down a little chatty and lengthy for providing simpler cases and motivation. $\endgroup$ – H. R. Nov 6 '17 at 22:51
  • $\begingroup$ @Rahul: I don't know when and why I got the -1!? At least, my answer is not wrong! :D $\endgroup$ – H. R. Nov 6 '17 at 22:55

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