Well, my definition of span is the following:


Suppose a vector space $(V,+,\cdot)$, and

$$S = \{u_1,\cdots,u_n\}$$

(and $S$ is a subset of $V$, not a subspace)

$$[S]=:\cap_{w\subset V, w\supseteq S} W$$

In other words, $[S]$ is, by definition, the intersection of all $W$, such that $W$ is a subset of $V$ and $W$ contains $S$.

So, if I want to find $[\emptyset]$, I need to look for all the subspaces that contains $\emptyset$. Well, every subspace contains $\emptyset$, because every subspace is a set. Therefore, I could imagine all diferente types of subspaces. Their intersection surely is gonna have the $0$ vector, because every subspace must contain it. But what guarantees that the intersection of the definition is gonna have only the zero vector? Agaun: I can imagine a lot of subspaces that contain $S$ (in this case, $\emptyset$), but also a lot of other elements. The definition says nothing about the space $V$.

I've found this proof, but I can't understand where did he took "smallest subspace possible", from the definition.

Also, in which way can we find the "basis for the empty set"? Is this possible? My teacher said to us that we should find it. Is the basis for the empty set, the $0$ vector of $V$?

  • 1
    $\begingroup$ $\{0\}$ is a subspace itself containing $\emptyset$, so $[S]\subseteq\{0\}$. Conversely, $0\in W$ for all subspaces $W$, so $\{0\}\subseteq[S]$. This shows the equality. $\endgroup$
    – Matt
    Dec 2, 2014 at 23:22

1 Answer 1


Note that the zero subspace, which is simply the set $\{0\}$, is one of the subspaces in your intersection and hence that intersection cannot have any vectors in it other than $0$.

Incidentally, as you have already noted that every subspace must contain $0$, the zero subspace, which only contains $0$, is the "smallest possible subspace", as Henno points out in your link.

  • $\begingroup$ Why the intersection can't have more than the $0$ vector? I can imagine the set $\{0\}$ inside the intersection, but the intersection being of sets such that they also have another elements in it $\endgroup$ Dec 2, 2014 at 23:51
  • $\begingroup$ Also, what would be the basis for the empty set? The $0$ vector? Because it can generate $\{0\}$? $\endgroup$ Dec 2, 2014 at 23:52
  • $\begingroup$ An intersection is by definition the elements that are contained in all the sets being intersected, so the elements of the intersection must be contained in $\{0\}$. $\endgroup$
    – Jim
    Dec 2, 2014 at 23:56
  • 1
    $\begingroup$ The set $\{0\}$ is not linearly independent so it is not a basis. The empty set $\emptyset$ is a basis for the zero subspace. $\endgroup$
    – Jim
    Dec 2, 2014 at 23:58
  • $\begingroup$ Thank you, I finally understood it. Do you have a book that Works with this definition of spam? I can only find the ones that uses the summatory definition $\endgroup$ Dec 3, 2014 at 1:19

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