Page 68 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes
and v w 1 w 2 v w 1 and v w since w is subspace, hence
1
2
u
, F and ,v w u v w and with the same argument
1 1
, F and ,v w 2 u v w 2 .
u
Therefore u v w and u v w .
1 2
u v w w .
1 2
Thus w w is a subspace of V F .
1 2
Theorem 2: The union of two subspaces is a subspace if one is contained in the other.

Proof: Let W 1 and W 2 be two subspaces of a vector space V.
Let W 1 W 2 or W 2 W 1 . Then W 1 W 2 or W (whichever is the case). Since W 1 ,W are subspaces
1
2
of ,V W 1 W is also a subspace of V.
2
Conversely, suppose W W is a subspace of V then we have to prove W W or W W .
1 2 1 2 2 1
Suppose it is not so, i.e., let us assume that W is not a subset of W and W is also not a subset
1 2 2
of W .
1
If W is not a subset of W then it implies that there exists
1
2
W 1 and W 2 ...(i)

Similarly if W is not a subset of W then there exists
2 1
W and W ...(ii)
2 1
From (i) and (ii) we see that

W 1 W 2 and W 1 W 2 since W 1 W is a subspace of ,V W 1 W 2
2
But W W W or W .
1 2 1 2
Suppose it belongs to W then since W and W is a subspace of ,V W which is
1 1 1 1
contradiction. Similar contradiction is arrived by assuming W 2 .
Therefore, either W W or W W .
1 2 2 1

2.2 Illustrative Examples

Example 1: Prove that the set W of ordered tried (a , a , 0) where
1 2
a ,a F is a subspace of V F , ,
1 2 3
Solution: Let a a 1 ,a 2 ,0 and b b 1 ,b 2 ,0 be two elements of W.

b
a
Therefore a 1 , , ,b 2 F . Let ,b F then
a
2
1

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