# Subspace proof, just need a little push

• Sep 30th 2008, 04:08 AM
Jameson
Subspace proof, just need a little push
Let $\displaystyle W_1 , W_2$ be non-empty subspaces of V. Prove that if $\displaystyle W_1 \cup W_2$ is a subspace of V, than either (I don't know how to do this in Latex) $\displaystyle W_1 \le W_2$ or $\displaystyle W_2 \le W_1$. ( $\displaystyle \le$ symbol should mean proper subset here).

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

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

Now I'm having a brain fade. Hint perhaps?
• Sep 30th 2008, 06:11 AM
bobak
Quote:

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

Let $\displaystyle V=R^3$...

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

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

Bobak
• Sep 30th 2008, 06:16 AM
Jameson
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?
• Sep 30th 2008, 06:44 AM
PaulRS
Suppose,$\displaystyle W_1 \not\subset W_2$ and $\displaystyle W_2 \not\subset W_1$ then there's a vector, call it $\displaystyle v_1 \in W_1 /v_1 \notin W_2$

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

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

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

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

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

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