How does just closure property of addition & scalar multiplication for a subset W of vector space V satisfies other axioms of vector spaces for W?

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.

• You ought to note that what you say is False. You also need the hypothesis $W\ne\emptyset$. Aug 10 '21 at 19:23
• @ancientmathematician Edited. Aug 11 '21 at 3:01

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 $$a$$).
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 $$a+b=0$$.