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.
0 comments:
Post a Comment