Statement:- The necessary & sufficient conditions for a non-empty subset W of a
vector spaceV(F) to be a subspace of V are
(
1) α∊W, β∊W ⇨ α-β∊W
(2) a∊F, α∊W ⇨ aα∊W
Proof:- Necessary Condition:-
Let V be
a vector space over the field F and W be its subspace.
∴
W will be a vector space over the same field F.
(W,+) is an abelian group.
∴
If β∊W ⇒ -β∊W
α∊W, -β∊W ⇒ α+(-β)∊W (By inverse axiom)
⇒ α-β∊W
Hence,
α∊W, β∊W ⇒ α-β∊W ,∀ α,β∊W
Also,
W will be closed under scalar multiplication.
∴
If a∊F, α∊W ⇒ aα∊W , ∀ a∊F, α∊W
Sufficient Condition:-
Let W be
a non-empty subset of a vector space V(F) satisfying the condition-
1- α∊W, β∊W ⇒ α-β∊W, ∀ α,β∊W
2- a∊F, α∊W ⇒ aα∊W, ∀ a∊F, α∊W
∴ from(1)
α∊W, β∊W ⇒ α-β∊W
⇒ α∊W, α∊W ⇒ α-α∊W
i.e. 0∊W
∴ The zero vector of V also exists inW.
0∊W, β∊W ⇒ 0-β∊W
⇒ -β∊W
∴ β∊W ⇒ -β∊W, ∀ β∊W
Hence,
each element of W have their additive inverse.
∵ α∊W, -β∊W
⇒ α-(-β)∊W
⇒
α+β∊W
W is
closed under vector addition.
∴
from(2)
a∊F,
α∊W ⇒ aα∊W, ∀ a∊F, α∊W
i.e. W is closed under scalar multiplication.
Since,
All the elements of W are the elements
of V. So associativity and community of vector addition are satisfied by
elements of W.
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