Union of subspaces: proving a biconditional statement

[thelema418]

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

 

[verty]

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

 

[thelema418]

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]

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 $x_1 + x_2 = z \in W$.

 

[thelema418]

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.

 

[bloby]

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

 

[Quesadilla]

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]

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?

 

[thelema418]

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 $V$.