# Proof involving vector subspaces

1. Sep 8, 2011

### strman

Let W1 and W2 be subspaces of a vector space V. Prove that W1$\bigcup$W2 is a subspace of V if and only if W1$\subseteq$W2 or W2$\subseteq$W1

Well so far, I have proven half of the statement (starting with the latter conditions). Right now I'm struggling to show that the final conditions follow from W1$\bigcup$W2. I have an idea for the method: assume that W1$\subseteq$W2 is not true, and then prove that W2$\subseteq$W1 must follow.

This is my first linear algebra class, and this is from the first problem set due, so I know that nothing that complex is going on here. However, I've been looking at this problem for the past 30 minutes or so, and I'm hoping that someone could push me in the right direction.

2. Sep 8, 2011

### lanedance

how about directly doing it? needs a bit of flesh...

W1 U W2 is subspace
take w1 in W1, w2 in W2
w1+w2 is in W1 U W2
but this means w1+w2 is in either W1 or W2
hence either w1 is in W2 or w2 is in W1

The problem with the way you describe might be if W1⊆W2 is not true, then this excludes W1=W2, which means you can only prove a proper subset, but i need to think about it more

3. Sep 8, 2011

### HallsofIvy

Staff Emeritus
I would use an "indirect proof". suppose it is NOT true that either W1 is a subset of W2 or W2 is a subset of W1. Then there exist a vector w1 that is not in W2 and a vector w2 that is not in W1. If w1+ w2= v is in W1 union W2, it is in either W1 or W2. But w1= v- w2, etc.

It might help to look at a simple example. Suppose W1 is the x-axis (as a subspace of the plane) and W2 is the y- axis. Are there sums of things in W1 and W2 that are not in their union? Generalize that idea.