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Let $W\subseteq V$ and $V$ is a vector space over a field $F=\mathbb{R }$.

I have read that if $W$ is closed under given addition and scalar multiplication, it will automatically satisfy the other axioms of vector space and hence then can be called a subspace of $V$, given $W$ is not an empty set. I understood all other axioms except i) existence of additive identity ii) existence of additive inverse.

Consider, the operations of vector addition and scalar multiplication on $V$ is not usual addition or usual scalar multiplication.

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    $\begingroup$ You ought to note that what you say is False. You also need the hypothesis $W\ne\emptyset$. $\endgroup$ Aug 10, 2021 at 19:23
  • $\begingroup$ @ancientmathematician Edited. $\endgroup$
    – novice
    Aug 11, 2021 at 3:01

2 Answers 2

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If $W$ is closed under addition and scalar-multiplication, we have mappings $+:W\times W\rightarrow W$ (linear mapping) and $*:R\times W\rightarrow W$ (scalar mult.) which are restrictions of the corresponding mappings for $V$. Since $V$ is a vector space containing $W$, the vector-space axioms are also fulfilled for the elements of $W$.

The situation is not so easy. In a group $G$ (such as the additive group of vectors of a vector space), a nonempty subset $U$ of $G$ forms a subgroup if $U$ is closed under the addition of group elements (group operation), so with $a,b\in U$ also $a+b\in U$, and for each group element $a\in U$ also $-a$ lies in $U$ (additive inverse of $a$).

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This is obvious. For an additive identity: Since $V$ is a vector space there exists $0_V\in V$ with $$x+0_V=x\quad(x\in V).$$Since every element of W is in $V$ this contains the fact that $$x+0_V=x\quad(x\in W),$$so $0_V$ is an additive identity for W$.

Similarly for additive inverses: Say $a\in W$. Then $a\in V$, so there exists $b\in V$ with $a+b=0$. Hence there exists $b\in W$ with $a+b=0$.

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