# Union of subspaces: proving a biconditional statement

## Homework Statement

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##.

## Homework 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##.

## 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

#### Attachments

• nondisjointsets.png
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## Answers and Replies

verty
Homework Helper
You've tried to combine the zero element with closure of scalar multiplication but have succeeded. But have you tried closure of addition?

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.

pasmith
Homework Helper
Try the contrapositive of $P \implies Q$: $\neg Q \implies \neg P$.

Suppose $W_1$ and $W_2$ are not subsets of each other, so that both $W_1 \setminus (W_1 \cap W_2)$ and $W_2 \setminus (W_1 \cap W_2)$ are non-empty. Let $x_1 \in W_1 \setminus (W_1 \cap W_2)$ and $x_2 \in W_2 \setminus (W_1 \cap W_2)$ and consider $x_1 + x_2 = z \in W$.

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.

For intuition it's better to think of subspaces of ##R^3## (lines, plans through the origin) than the sketch you draw

Try the contrapositive of $P \implies Q$: $\neg Q \implies \neg P$.

Suppose $W_1$ and $W_2$ are not subsets of each other, so that both $W_1 \setminus (W_1 \cap W_2)$ and $W_2 \setminus (W_1 \cap W_2)$ are non-empty. Let $x_1 \in W_1 \setminus (W_1 \cap W_2)$ and $x_2 \in W_2 \setminus (W_1 \cap W_2)$ and consider $x_1 + x_2 = z \in W$.

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?

pasmith
Homework Helper
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.

If $z \in W_1 \cup W_2$ then $z \in W_1$ or $z \in W_2$. Is either of those possible, bearing in mind that if two vectors are members of a subspace then so is their difference?

1 person
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##.