Subspace proof, just need a little push

**Jameson** · September 30th 2008, 04:08 AM

Let $W_1 , W_2$ be non-empty subspaces of V. Prove that if $W_1 \cup W_2$ is a subspace of V, than either (I don't know how to do this in Latex) $W_1 \le W_2$ or $W_2 \le W_1$ . ( $\le$ symbol should mean proper subset here).

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This should be done by contradiction I think. So assume that there exists $x_1 \in W_1$ and $x_2 \in W_2$ , wuch that $x_1 , x_2 \notin W_1 \cap W_2$ .

Still assume now that $W_1 \cup W_2$ is a subspace of V.

Now I'm having a brain fade. Hint perhaps?

**bobak** · September 30th 2008, 06:11 AM

Originally Posted by hatsoff

I don't think that is true (and thus can't be proven). I'm rusty on subspaces but...

Let $V=R^3$ ...

Let $W_1$ be the set spanned by $\left[\begin{array}{c}0\\0\\1\end{array}\right]$ . Let $W_2$ be the set spanned by $\left[\begin{array}{c}0\\1\\0\end{array}\right]$ . Then $W_1\cup W_2$ is the set spanned by $\left[\begin{array}{c}0\\1\\1\end{array}\right]$ . So $W_1$ , $W_2$ and $W_1\cup W_2$ are subspaces of $V$ , but $W_1$ and $W_2$ are non-overlapping.

The vector O must be a member of both subspaces so they will overlap. Also you have defined $W_1\cup W_2$ incorrectly, what you have given is the sum of $W_1$ and $W_2$ . the union is not a subspace.

Bobak

**Jameson** · September 30th 2008, 06:16 AM

So bobak, what's your take on this problem?

Now is overlapping what we don't want here? For instance, if the zero vector is in both subspaces or any vector in both, does the union and overlap take away the uniqueness of terms?

**PaulRS** · September 30th 2008, 06:44 AM

Suppose, $ W_1 \not\subset W_2 $ and $ W_2 \not\subset W_1 $ then there's a vector, call it $ v_1 \in W_1 /v_1 \notin W_2 $

There's also $ v_2 \in W_2 /v_2 \notin W_1 $

We have $ v_1 ,v_2 \in \left( {W_1 \cup W_2 } \right) $ so if the union is a subspace we should get $ v_1 + v_2 \in \left( {W_1 \cup W_2 } \right) $

However $ v_1 + v_2 \notin W_1 $ and $ v_1 + v_2 \notin W_2 $ since $ v_1 + v_2 \in W_1 $ would imply ( by definition of subspace, summing $-v_1$ ) that $ v_2 \in W_1 $ analog. with $ v_1 + v_2 \notin W_2 $

So $ v_1+v_2 \notin \left( {W_1 \cup W_2 } \right) $ absurd

So our assumption $ W_1 \not\subset W_2 $ and $ W_2 \not\subset W_1 $ , must have been wrong

If one is contained in the other, the union is obviously a subspace, and so we are done

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