# Question 2

Note that if U and W are both vector spaces, then they both contain the zero vector for V. So U \setminus W does not contain the zero vector for V and so cannot form a subspace.

Note that if U \cup W is a subspace of V then the zero vector for V is contained in both U and W (obviously satisfied) and x + \lambda y \in U \cup W for all x, y \in U \cup W. Note that if both x \in U and y \in U (or similarly x \in W and y \in W), then x + \lambda y \in U \cup W since U, W are subspaces.

We can guarantee that for all x, y \in U \cup W we have x, y \in U if W \subseteq U, so that U \cup W = U. Similarly, we can guarantee that for all x, y \in U \cup W we have x, y \in W if U \subseteq W so that U \cup W = W. So if either U \subseteq W or W \subseteq U we have U \cup W is a subspace. So we’ve found a sufficient condition.

The problem comes when, say, x is only in U and y is only in W or vice versa. Could x + \lambda y be in U \cup W? Turns out not.

We can show that our sufficient condition is also a necessary condition. We show that if U \not \subseteq W and W \not \subseteq U then U \cup W is not a subspace. As U \not \subseteq W there exists an x \in U with x \not \in W and as W \not \subseteq U, there exists a y \in U with y \not \in W. If U \cup W were to be a subspace, we would have x + y \in U \cup W, ie. either x + y \in U, x + y \in W or both. We have:

• If x + y \in U, since \pm x \in U, we have x + y - x = y \in U, contradiction.
• If x + y \in W, since \pm y \in W we have x + y - y = x \in W, contradiction.

Both contradictions arising since we picked x, y such that they are not in both U and W.

So if U \cup W is a subspace then U \subseteq W or W \subseteq U.

Overall - we have that U \cup W is a subspace iff U \subseteq W or W \subseteq U.

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