- #1
Devil Moo
- 44
- 1
This is a linear algebra question which I am confused.
1. Homework Statement
Prove that "if the union of two subspaces of ##V## is a subspace of ##V##, then one of the subspaces is contained in the other".
Suppose ##U##, ##W## are subspaces of ##V##. ##U \cup W## is a subspace of ##V##. (statement A)
Suppose ##U## is not contained in ##W## and vice versa. (statement B)
Let ##u \in U, \not\in W## and ##w \in W, \not\in U##.
##u \in U \cup W, w \in U \cup W##
##u + w \in U \cup W##
##u + w \in U## or ##u + w \in W##
Suppose ##u + w \in U## (statement C)
##u + w + (-u) \in U##
##w \in U##
It leads to contradiction.
Which one does it conclude? (if A is true, then B is false) or (if A and B are true, then C is false)
1. Homework Statement
Prove that "if the union of two subspaces of ##V## is a subspace of ##V##, then one of the subspaces is contained in the other".
The Attempt at a Solution
Suppose ##U##, ##W## are subspaces of ##V##. ##U \cup W## is a subspace of ##V##. (statement A)
Suppose ##U## is not contained in ##W## and vice versa. (statement B)
Let ##u \in U, \not\in W## and ##w \in W, \not\in U##.
##u \in U \cup W, w \in U \cup W##
##u + w \in U \cup W##
##u + w \in U## or ##u + w \in W##
Suppose ##u + w \in U## (statement C)
##u + w + (-u) \in U##
##w \in U##
It leads to contradiction.
Which one does it conclude? (if A is true, then B is false) or (if A and B are true, then C is false)