Statement:-
The union
of two subspace of a vector space is a subspace iff one each contained.
Proof:-
Let V be
a vector space over the field F.Let W1 and W2 be its subspaces
such that either
If W1⊆W2 then
W1∪W2=W2
Since, W2 is a subspace.
Hence W1∪W2 is also a subspace.
Again, if W2⊆W1
then W1∪W2=W1
Since, W1 being a subspace.
Hence W1∪W2 will also be asubspace.
Conversally:-
Let W1
and W2 be subspaces of V such that W1∪W2 be also a subspace.
Let if
possible W1⊈W2 or W2⊈W1
Since if
W1⊈W2
⇨ there exists αєW1 and α∉W2
But α∊W1∪W2
If W2⊈W1
⇨ there exists β∊W2 and β∉W1
But β∊W1∪W2
Hence W1∪W2 is a subspace.
Since α∊W1∪W2 ,
β∊W1∪W2
⇨ α+β∊W1 or
α+β∊W2
Let α+β∊W1
Hence α+β∊W1 and α∊W1
⇨
(α+β)-α∊W1
⇨
β∊W2
Which is contradiction that β∉W1
Let α+β∉W2
Since α+β∊W2 and β∊W2
⇨ (α+β)-β∊W2
⇨ α∊W2
Which is contradiction that α∉W2
Hence W1⊆W2 or W2⊆W1 Proved
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