Union of Subspaces of V: Proving Containment

[Awatarn]

Prove that the union of two subspaces of V is a subspace of V if and only if one of the subspaces is contained in the other.

 

[matt grime]

Yep, what hav you done to try and figure out the answer? Have you tried both directions? The 'if' part is straightfoward. You might want to try contradiction for the 'only if' part.

 

[Awatarn]

I. The straight forward part:
1. for U_2, U_1 are subspaces of V
2. let U_1 is contained in U_2.
\therefore U_1 \cup U_2 = U_2
\therefore U_1 \cup U_2 = U_2is also subspaces ofV

II. the 'only if' part:
1. For any U_1, U_2 are subspace of V, they must contain U_1 \oplus U_2 which is the smallest subspaces containing in U_1, U_2
2. therefore if there is w_1 \notin U_1 ; w_1 \in U_1 \cup U_2, it will contradict to the statement 1.
3. the only way of existing of w_1 is that U_2 \cup U_1 = U_1 or U_2 is contained in U_1

The proove finished. Is there sufficeintly complete?

 

[mathwonk]

note the union of two distinct lines is not closed under forming poarallelograms

 

[matt grime]

The second part is not correct. Just show that the union of subspaces does not give a subspace; think about mathwonk's hint. What must a subspace satisfy? Closure under addition.

 

[Awatarn]

I edit the second statement of the seconde part to:
2. Give w_1 \notin U_1 ; w_1\in U_1 \cup U_2. If they will form subspace, it must write their linear combination in form of
au_1 + bw_1 where u_1 \in U_1 and w_1 \in W.
This linear combination have not closure under addition, if w_1 is not contain in U_1

Is it OK?

 

[matt grime]

No, not in my opinion- you've just asserted the result is true without explaining why. Now, I understand why it is true, but it does not convince me that you understand why it is true, which is what you're really attempting to show.

 

[Awatarn]

2. Give w_1 \notin U_1 ; w_1\in U_1 \cup U_2. If they will form subspace, it must write their linear combination in form of
au_1 + bw_1 where u_1 \in U_1 and w_1 \in W.
This linear combination have not closure under addition, if w_1 is not contain in U_1

According to mathwonk's idea, u_1 \in U where it is the line in 3D and w_1 \in W where it is the line in 2D. U and W is not the same line in 2D. U may be the line of y=1 and W is the line of y=2 When we write the linear combination of au_1 + bw_1, it will form a new line.U may be the line of y=1 and W is the line of y=2. Their linear combination will form line of y=3 . therefore the linear combination have not closure under addition.