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My book says that if it is given that a set is in a vector space then the only properties that need to be shown are closure under addition and scalar multiplication, existence of a zero vector, and existence of an additive inverse. I understand why the all the rest need to be shown or not shown, except for the additive inverse. The only way I can think of it being violated is if the zero vector doesn't exist inside the subspace. But if that's the case then why show this property in addition to the zero vector? Is there some other way for it to fail?

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Closure under addition and scalar multiplication is enough, since the additive inverse and the zero vector may be obtained by multiplying vectors with the scalars $-1$ and $0$, respectively. You do have to make sure the set is not empty, though, and the easiest vector to check whether it's actually in there is usually the zero vector.

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  • $\begingroup$ Thank you! That is a much simpler way of viewing it. I very much appreciate the help. $\endgroup$ – Waffles Aug 2 '17 at 18:39
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Your complaint is valid.

Existence of additive inverses are guaranteed via scalar multiplication by the field element $-1$.

In fact, as long as the given set is nonempty, you don't even need to show the existence of a zero vector, since that's implied via scalar multiplication by the field element $0$.

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In fact, if a subset $A$ of a vector space $V$ is non-empty, closed under addition and scalar multiplication, then it is already a vector space in its own right. In fact it must contain zero because it is non-empty, so there is some $x\in A$, and it is closed under multiplication so $-x\in A$, and it is closed under addition, so $x+(-x)\in A$. So you are correct in your observations that in order to check that a subset of a vector space is itself a vector space, you need not go through the entire list of axioms of a vector-space, but only check that it is closed under addition and scalar mutliplication.

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