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Union of subspaces: proving a biconditional statement

  1. Jun 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Let ##W_1## and ##W_2## be subspaces of a vector space ##V##. Prove that ##W_1 \cup W_2## is a subspace of ##V## if and only if ##W_1 \subseteq W_2## or ##W_2 \subseteq W_1##.

    2. Relevant equations

    A subset ##W## of a vector space ##V## is a subspace of ##V## provided that (a) ##W## is closed under addition, (b) ##W## is closed under scalar multiplication for all ##c \in F##, and (c) the zero element of ##V## is the zero element of ##W##.

    The problem statement is biconditional. Therefore, we need to show both ##P \Rightarrow Q## and ##Q \Rightarrow P##.

    3. The attempt at a solution

    Demonstration that ##Q \Rightarrow P##. Let ##W_1## and ##W_2## be subspaces of ##V##. Consider ##W_1 \subseteq W_2## (or ##W_2 \subseteq W_1##). If ##W_1 \subseteq W_2##, then ##W_1 \cup W_2 = W_2##. Since ##W_2## is a subspace, ##W_1 \cup W_2## is a subspace by equality. A similar argument can be presented for ##W_2 \subseteq W_1 = W_1##. Therefore, if ##W_1 \subseteq W_2## or ##W_2 \subseteq W_1##, then ##W_1 \cup W_2## is a subspace of ##V##.

    Incomplete demonstration ##P \Rightarrow Q##. Let ##W_1## and ##W_2## be subspaces of ##V##. Suppose ##W_1 \cup W_2## is a subspace of ##V##. By definition of subspaces, ##W_1## and ##W_2## contain the same zero element from ##V##. Therefore, ##W_1 \cap W_2 \neq \varnothing##: this means that the sets are not disjoint.

    By definition of subspaces, scalar multiplication is closed. There exists ##c(x_1) \in W_1 \cup W_2## for every ##x_1 \in W_1 \cup W_2## and ##c \in F##. (...)

    I'm not really certain how to develop this last part. I chose to work with the condition of scalar multiplication because ##c \in F## has to be the same in each subspace. I attached a picture of the situation I have. I need to show that only the subset situation follows.

    Thanks
     

    Attached Files:

  2. jcsd
  3. Jun 19, 2014 #2

    verty

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    You've tried to combine the zero element with closure of scalar multiplication but have succeeded. But have you tried closure of addition?
     
  4. Jun 19, 2014 #3
    I did not succeed with the closure of scalar multiplication. I need to show that one set must be a subset of the other.

    I don't see how the closure of addition helps. If I take two elements from the joined sets then their addition has to be within the joined set. The only thing I know definitively is that the zero element is in both the sets. If I had a second element that was in both sets, then I could do a lot more with addition. Yet, I don't see how P leads me to that conclusion.

    I also tried a proof by contradiction, but that also gets me into this same situation.
     
  5. Jun 19, 2014 #4

    pasmith

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    Try the contrapositive of [itex]P \implies Q[/itex]: [itex]\neg Q \implies \neg P[/itex].

    Suppose [itex]W_1[/itex] and [itex]W_2[/itex] are not subsets of each other, so that both [itex]W_1 \setminus (W_1 \cap W_2)[/itex] and [itex]W_2 \setminus (W_1 \cap W_2)[/itex] are non-empty. Let [itex]x_1 \in W_1 \setminus (W_1 \cap W_2)[/itex] and [itex]x_2 \in W_2 \setminus (W_1 \cap W_2)[/itex] and consider [itex]x_1 + x_2 = z \in W[/itex].
     
  6. Jun 19, 2014 #5
    I think this is still the same problem I'm having with my other methods.

    Here ##\neg P## would be "##W_1 \cup W_2## is not a subspace of V". I don't see how ##x_1 + x_2 = z \in W## presents a problem with the closure of addition on ##W_1 \cup W_2##.

    So, whatever I'm not seeing in this, is the same issue I'm not seeing in the other methods I've been trying.
     
  7. Jun 20, 2014 #6
    For intuition it's better to think of subspaces of ##R^3## (lines, plans through the origin) than the sketch you draw
     
  8. Jun 20, 2014 #7
    Continue from this. ##x_1,x_2 \in W_1 \cup W_2## but does ##x_1 + x_2 \in W_1 \cup W_2##? What would that mean? Can you reach a contradiction?
     
  9. Jun 20, 2014 #8

    pasmith

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    If [itex]z \in W_1 \cup W_2[/itex] then [itex]z \in W_1[/itex] or [itex]z \in W_2[/itex]. Is either of those possible, bearing in mind that if two vectors are members of a subspace then so is their difference?
     
  10. Jun 20, 2014 #9
    So... something like this: Showing ##\neg Q \Rightarrow \neg P##. Suppose that the subspaces ##W_1## and ##W_2## are not subsets of each other. Let ##x_1 \in W_1## with ##x_1 \notin W_2##, and let ##x_2 \in W_2## with ##x_2 \notin W_1##. (Observe that ##x_1 \neq 0## and ##x_2 \neq 0##.) By definition of a subspace, addition must be closed for all elements of the set. Then ##x_1 + x_2 = z \in W_1 \cup W_2##. This means that ##z## must be a member of ##W_1## or ##W_2## (or both).

    Consider the case where ##z \in W_1##. By definition of a vector space, ##x_1## has an additive inverse, denoted ##-x_1##, and this is also an element of ##W_1 \cup W_2##. Then ## -x_1 + z \in W_1##. However, ##-x_1 + z = -x_1 + (x_1 + x_2) = (-x_1 + x_1) + x_2 = 0 + x_2 = x_2 \notin W_1##.

    Now consider the case where ##z \in W_2##. By definition of a vector space, ##x_2## has an additive inverse, denoted ##-x_2##, and this is also an element of ##W_1 \cup W_2##. Then ## -x_2 + z \in W_2##. However, ##z + -x_2 = (x_1 + x_2)+ -x_2 = x_1 + (x_2 + -x_2)= x_1 + 0 = x_1 \notin W_2##.

    Consequently, addition is not closed, and ##W_1 \cup W_2## is not a subspace of ##V##.
     
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