Theorem-Direct Sum

Statement:- The necessary and sufficient condition for a vector space V over the field F to be direct sum of its two subspaces W₁ and W₂ are that

1-V=W₁+W₂

2- W₁ and W₂ are disjoint i.e.  W₁∩W₂={0}

Proof:-       Necessary Condition:-

Let V be a vector space over the field F and W1,W2 be its subspaces.  Let V be direct sum of W1 and W2. i.e.

⇨               each element of V is expressible uniquely as a sum of an

element   of W1 and an element of W2.

In particular each element of V is expressible as a linear sum of an element of W1 and an element of W2.

i.e.                        V=W1+W2

Let                       0≠W1∩W2 also αєV

∴                        we can write

α=α+0,   α∊W1,  0∊W2

α=0+α,   0∊W1 & α∊W2

This shows that a non-zero vector α∊V is expressible in at least two different way as sum of an element of W1 an element Of W2.

Which is contradiction that V is direct sum of W1 and W2.

Thus only zero vector common to both W1 and W2.

∴                      W1∩W2={0}

Sufficient Condition:-

Let V be a vector space and W1,W2 be its subspaces satisfying the condition-

1-                           V=W1+W2       …………………………..(1)

2-                            W1∩W2={0}  ……………………………(2)

therefore       from(1)

Each element of V is expressible as a sum of an element of W1 and an element of W2.

Let              0≠α∊V be expand as

such that   α=α1+β1,for some  α1∊W1 and β1∊W2

and             α=α2+β2, for some α2∊W₁ and β₂∊W₂

⇨               α₁+β₁=α₂+β₂

or               α₁-α₂=β₂-β₁         …………………………..(3)

Since         α₁,α₂∊W₁ ⇨ α₂-α₁∊W₁

also,          β₁,β₂∊W₂ ⇨  β₂-β₁∊W₂

Therefore from(3)

α₁-α₂=β₂-β₁∊W₁∩W₂

But from (2)

W₁∩W₂={0}

⇨                  α₁-α₂=0 ⇨  α₁=α₂

And               β₂-β₁=0 ⇨  β₁=β₂

Thus each element of V is expressed uniquely as a sum of an element W₁ and an element of W₂.

∴                    V=W₁⊕W₂.