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Question

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 from which this answer can be deduced.

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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
    Commented Aug 31, 2016 at 11:53
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    $\begingroup$ @H.R.: Related: math.stackexchange.com/questions/1812653 $\endgroup$
    – Watson
    Commented Aug 31, 2016 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
    Commented Aug 31, 2016 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
    Commented Aug 31, 2016 at 13:12
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    $\begingroup$ @H.R. math.stackexchange.com/questions/734418/… $\endgroup$
    – IAmNoOne
    Commented Aug 31, 2016 at 13:21

3 Answers 3

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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
    Commented Aug 31, 2016 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
    Commented Aug 31, 2016 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
    Commented Aug 31, 2016 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
    Commented Sep 1, 2016 at 8:55
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    $\begingroup$ @H.R. No, in fact, it should say "finite non-empty". $\endgroup$
    – 5xum
    Commented Sep 1, 2016 at 9:29
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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
    Commented Aug 31, 2016 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
    Commented Aug 31, 2016 at 12:35
  • $\begingroup$ So the OP's question ultimately depends on the definition of span. $\endgroup$
    – Fei Li
    Commented Aug 31, 2016 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
    Commented Sep 1, 2016 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
    Commented Sep 1, 2016 at 11:51
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A basis has several equivalent definitions. One of which is:

  1. A basis of a vector space is a minimal generating set

So keeping that in mind, if we look at $V = \{0\}$, the only non-empty subset of this vector space is $B = \{0\}$. This set $B$ is a linearly dependent set and thus it cannot be a basis.We make a note here that $B$ is a generating set of $V$. We know by existence of basis of a vector space that $V$ must also have a basis. So if a basis were to exist it should be a subset of $B$. As removing a linearly dependent element from a generating set does not change the span of that set, $\phi$ is a generating set.

Now the definition of a linearly dependent set in crude language is: In a vector space $V$, a subset $A$ of $V$ is said to be linearly dependent if there exists an element which can be written as a finite linear combination of the rest of the elements.

So, consider the set $\phi$; there does not exist any element in it which can be written as a finite linear combination of the other elements, hence it is not a linearly dependent set and therefore it is linearly independent.

Therefor we see that $\phi$ is a linearly independent and generates $V$ and hence a basis.

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    $\begingroup$ $\emptyset$ should be used for empty set, not $\phi$ , or you can use $\varnothing$ $\endgroup$
    – john
    Commented Jun 4, 2021 at 4:28

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