Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

About vector space1

  1. Jun 7, 2006 #1
    Prove that the union of two subspaces of [tex]V[/tex] is a subspace of [tex]V[/tex] if and only if one of the subspaces is contained in the other. :approve:
     
  2. jcsd
  3. Jun 7, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Jun 7, 2006 #3
    I. The straight forward part:
    1. for [tex]U_2, U_1 [/tex] are subspaces of [tex]V[/tex]
    2. let [tex]U_1[/tex] is contained in [tex]U_2[/tex].
    [tex]\therefore U_1 \cup U_2 = U_2[/tex]
    [tex]\therefore U_1 \cup U_2 = U_2 [/tex]is also subspaces of[tex] V[/tex]

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

    The proove finished. Is there sufficeintly complete? :uhh:
     
    Last edited: Jun 7, 2006
  5. Jun 7, 2006 #4

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    note the union of two distinct lines is not closed under forming poarallelograms
     
  6. Jun 8, 2006 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  7. Jun 8, 2006 #6
    I edit the second statement of the seconde part to:
    2. Give [tex]w_1 \notin U_1 ; w_1\in U_1 \cup U_2 [/tex]. If they will form subspace, it must write their linear combination in form of
    [tex]au_1 + bw_1[/tex] where [tex]u_1 \in U_1 and w_1 \in W[/tex].
    This linear combination have not closure under addition, if [tex]w_1[/tex] is not contain in [tex]U_1[/tex]

    Is it OK?
     
  8. Jun 8, 2006 #7

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
    Last edited: Jun 8, 2006
  9. Jun 13, 2006 #8
    According to mathwonk's idea, [tex]u_1 \in U[/tex] where it is the line in 3D and [tex]w_1 \in W [/tex] where it is the line in 2D. [tex]U[/tex] and [tex]W[/tex] is not the same line in 2D. [tex]U[/tex] may be the line of [tex]y=1[/tex] and [tex]W[/tex] is the line of y=2 When we write the linear combination of [tex]au_1 + bw_1[/tex], it will form a new line.[tex]U[/tex] may be the line of [tex]y=1[/tex] and [tex]W[/tex] is the line of y=2. Their linear combination will form line of [tex]y=3[/tex] . therefore the linear combination have not closure under addition.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?