Theorem-Vector Space

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.