Veactor Space (Theorem-1)

Statement-The necessary and sufficient condition for a non-empty subset W of vector space V is that

a,bєF, α,βєW ⇨ aα+bβєW, ∀ a,bєF and α,βєW

Proof- Necessary condition-

Let W be a subspace of a vector space V over the field F.

∵ W is a vector space over the same field F. So W is a closed under scalar multiplication.

Let a,bєF and α,βєW

Since aєF,αєW ⇨ aαєW

and bєF,βєW ⇨ bβєW

Since (W,+) is a abelian group.

Hence aαєW, bβ∊W⇨aα+bβ∊W (By closure axiom)

Hence a,b∊F, α,β∊W ⇨ aα+bβ∊W

Sufficient Condition:-

Let W be a non-empty subset of vector space V(F) satisfying the condition-

a,bєF, α,β∊W⇨ aα+bβєW, ∀a,b∊F and α,β∊W

Also 1∊F Now taking a=1, b=1

1,1∊F, α,β∊W ⇨1.α+1.βєW ⇨ α+βєW

Since W is closed under vector addition.

Taking a=0, b=1

⇨ 0.α+(-1)β∊W

⇨ -β∊W

The additive inverse of each elements exist in W.

Taking a=0, b=0 ⇨ 0.α+0.β∊W

⇨ 0∊W

Thus zero vector of V exist in W.Also,

∵ W⊆V

So,By associativity and commutivity of vector addition hold in W as they hold in V.

Hence , (W,+) is an abelian group.

Again, Taking β=0

We have aα+b.0∊W ⇨ aα+0∊W ⇨ aα∊W

Hence, W is closed under scalar multiplication.

The remaining postulates of vector space will hold in W.They hold in V of which W is a subset.

Thus W is a vector space or W is a subspace.

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