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To show a set is a vector subspace, I see it´s necessary to prove

a) An addition property: if $x$ and $m$ are both elements of the set, then $x+m$ must also be an element of the set for it to be a vector subspace.

b) A scalar multiplication property: if $x$ is an element of the set, then $kx$ is also defined in the set, where $k$ is a scalar.

However, many people insist that it´s also necessary to prove that the neutral element, $0$ is also defined in the set for it to be a vector subspace, independent of the other two properties.

Isn´t this property, however, implied by a) and b) as follows?:

Given the additive property, $x-m=0$, where $x=m$

or by the scalar property, $kx=0$, where $k=0$

Asaf Karagila
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XXb8
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    Well, technically you would also need to show that there is some element of the space. The empty set is not a vector space, though it is still (vacuously) closed under scalar multiplication. – lulu Oct 22 '21 at 11:44
  • @lulu If it´s given that there is an element of the space, then could you address the question? – XXb8 Oct 22 '21 at 11:46
  • Within reason, then yes. You can deduce the existence of $\vec 0$ from everything else if you are careful. See, for instance, this duplicate. Not sure this is that helpful though...is the alternate set of axioms really easier to work with? – lulu Oct 22 '21 at 11:51
  • To stress: the existence of inverses has to be rephrased in a way which does not reference $\vec 0$. – lulu Oct 22 '21 at 12:06

3 Answers3

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You need to show that there is an element in your set.

The empty set satisfies the “addition property” as well as the “scalar multiplication property”, so you have to ensure the set is not empty.

Usually, asking for $0$ to be in the set is the first property to check, but just being not empty is sufficient, along with the other mentioned properties.

Indeed, if $x_0\in S$, then $(-1)x_0\in S$ (scalar multiplication property) and therefore $$ x_0+(-1)x_0\in S $$ But it follows from the vector space axioms that $x_0+(-1)x_0=0$, so we're done.

Perhaps more simply: the scalar multiplication property implies that $0x_0=0\in S$ (but you still need to have an element to start with).

egreg
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  • This answer makes sense. But in the case we know the set is not empty from the start, it would mean it's unecessary to prove $0$ is in the set, correct? Just proving the two properties I mentioned would suffice. – XXb8 Oct 22 '21 at 13:38
  • Also, you would need at least two elements in the set (at least $x_0$ and it´s additive inverse $x_0$ ) to show $0$ is in the set too, not just one element making the set non-empty – XXb8 Oct 22 '21 at 13:42
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    @XXb8 The existence of $x_0\in S$ implies that of $ax_0$ for every scalar $a$. But these might well be all the same elements in case $x_0=0$. But anyway, you just need to show one element exists. – egreg Oct 22 '21 at 14:07
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Those properties rely on the existence of an element $0$ if it where not to exist how would you argue what is the value of $x-x$?

Sebastian Cor
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It is sufficient too prove that

  1. For any u, v in the subset, u+ v is in the subset.
  2. For any v in the subset and scalar a, av is in the subset.

That is, that the subset is closed under vector addition and closed under scalar multiplication.

Since the set of scalars is a field, it contains a 0 element and,for any vector v, 0v is the 0 vector.

user247327
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