0
$\begingroup$

I'm reading Friedberg's linear algebra:

(1) From definition of both "linear combination" and "span": The subset of concern should be nonempty.

(2) From definition of both "linearly dependent" and "basis": The subset of concern is ... just a subset.

But why (1) put additional restriction on the subset concerned?or, maybe, why (2) didn't?

I know that since the basis of $\{0\}$ is $\{\}$, so (2) have to include all subset including the empty one.

And since that the definition of basis use the word "linearly independent", so the definition of "linearly dependent" also include all subset including the empty one.

And, to the definition of "linearly dependent": "A subset S of a vector space V is called linearly dependent if there exist a finite number of distinct vectors ..."

So this means that $0$ is not a finite number?(But I think it's finite) since $\{\}$ has no elements and it's not linearly dependent?

linear combination definition span definition linearly dependent definition basis definition

$\endgroup$
  • 3
    $\begingroup$ Please include the definitions you refer to, so we can work with you in case you're making a mistake in your conclusions (1), (2). Most of us do not have a copy of your text immediately in front of us. $\endgroup$ – amWhy Mar 3 '18 at 17:52
  • 3
    $\begingroup$ The definition of "span" explicitly includes the empty set! And no, there's no reason the set has to be nonempty in the definition of "independent". The definition says $S$ is dependent if there exist $v_1,\dots,v_k\in S$ such that etc. That makes perfect sense if $S$ is empty - in that case there do not exist such vectors $v_j$, so the definition shows precisely that the empty set is not dependent. $\endgroup$ – David C. Ullrich Mar 3 '18 at 18:01
  • $\begingroup$ @DavidC.Ullrich May I ask you the reason the word "nonempty" is used in the definitions? $\endgroup$ – Postal Model Mar 3 '18 at 18:43
  • 1
    $\begingroup$ The definition of "span" is split into two cases, just to clarify what the span of the empty set is. The word "nonempty" is not used in the definition of "independent". $\endgroup$ – David C. Ullrich Mar 3 '18 at 18:51
  • 1
    $\begingroup$ If I were in charge I'd leave the word out in the definition of linear combination. Of course that might lead to confusion... $\endgroup$ – David C. Ullrich Mar 3 '18 at 19:01
1
$\begingroup$

A subset $S$ of a vector space $V$ over a field $\mathbb{F}$ is called linearly dependent if there are vectors $v_1,\dots,v_n$ and scalars $a_1,\dots,a_n$ such that $$ \sum_{k=1}^n a_k v_k = 0. $$ As such, all linearly dependent sets must be nonempty.

On the other hand, if $v$ is any nonzero vector, and $a$ is any nonzero scalar, then $av$ is nonzero. Consequently, the only possible basis for $\{ 0\}$ is itself — but this set is clearly linearly dependent since $1\cdot 0 = 0$. If every vector space is to have a basis, we are left with no other choice but to define $\text{span}(\emptyset) =\{ 0 \}$ (note that the empty-set is vacuously linearly independent).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ May I ask you that the reason the word "nonempty" is used is to avoid that span($\phi$) would be $\{\}$? $\endgroup$ – Postal Model Mar 3 '18 at 18:29
  • 1
    $\begingroup$ Intuitively, it doesn’t make much sense to define the span of the empty set as the trivial subspace, but this was an aesthetic choice made by mathematicians to ensure that every subspace has a basis. $\endgroup$ – Pietro Paparella Mar 3 '18 at 18:31
  • 1
    $\begingroup$ It makes perfect sense, from either of at least two points of view. (i) Say the span of $S$ is the smallest subspace containing $S$. (ii) Regard the sum of no vectors at all as the zero vector. $\endgroup$ – David C. Ullrich Mar 3 '18 at 18:54
  • $\begingroup$ @DavidC.Ullrich certainly, but regarding the empty sum as zero is also an aesthetic choice, so every direction leads to convention. Furthermore, I think the other explanation is more accessible to the learner of mathematics. $\endgroup$ – Pietro Paparella Mar 3 '18 at 19:01
3
$\begingroup$

Regarding whether the span of the empty set "should" be defined to be $\{0\}$: There are at least two ways to look at this.

(i) Rephrase the definition to say the span of $S$ is the smallest subspace containing $S$. That's exactly the same as "the set of all linear combinations of elements of $S$" if $S$ is nonempty, but it applies equally well to the empty set.

(ii) Regard the sum of no vectors at all to be the zero vector. After all, what else should it be? Or consider the natural way to define the sum of a finite set:

def sum(S):
  s=0
  for v in S:
    s=s+v
  return s
| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

For a given non-emtpy subset of a vector space $\mathsf{V}$, the $\mathbf{0_\mathsf{V}}$ is always achievable by just making all the scalars $0$. But if we apply the empty set to the definition of "linear combination", the result is that $\mathbf{0_\mathsf{V}}$ is not a linear combination of vectors in $\{\}$, since we don't have any vector to do it. This new empty-set-included definition will make the empty set very special, which is not what we want.

On the other side, there is no problem about $\{\}$ in the definition of linearly independent, so we don't have to specifically exclude the empty set.

After all some part of math be like that is because people want it to be like that.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.