Vector space,subspace and Lin. $X$ is a vector space with a scalar field $K$, and let $U$ and $V$ be subspaces of $X$. 
1) Prove that $ U \cup V$ is a subspace of $X$ if and only if $U \subset V   $ or $V \subset U  $. 
2) Show that $Lin(U \cup V)= U + V$, where $Lin(S)$ is the subspace of $X$ generated by $S$.
$ U \cup V$ is a subspace of X when, for every pair of vectors $\vec{a},\vec{b} \in U \cup V $ we have $\vec{a} + \vec{b} \in   U \cup V$ and for any scalar $\alpha \in K$, we have  $ \alpha \cdot\vec{a}\in U \cup V  $. How to solve the problem?
 A: Let $\alpha \in K$, and $a, b \in U \cup V$.
First, assume $a \in U$. Since $U$ is a subspace by itself, we have $\alpha \cdot a \in U \subset U \cup V$ and therefore any scalar multiple of a vector in $U$ is in $U \cup V$. A similar argument shows that if $a \in V$, then $\alpha \cdot a \in U \cup V$.
The only problem may lie in the additive axiom, since for $a \in U, \ b \in V $, we might not have $a + b \in U \cup V$. 
1) (If part) If $U \subset V$, then $a, b \in U \cup V$ implies $a, b \in V$, for if $a$ or $b \in U$ then they are in $V$. In this case, and since $V$ is a subspace, we have $a + b \in V \subset U \cup V$. The case where $V \subset U$ is similar.
(Only if part) If $U \cup V$ is a subspace, let $a \in U$, $b \in V$. By assumption, we have $a + b \in U \cup V$. Assume first that $a + b \in U$. Then $b = (a + b) - b \in U$ because both $a + b$ and $b \in U$, and $U$ is a subspace. We conclude that every $b \in V$ is in $U$, i.e., $V \subset U$. The case where $a + b \in V$ is similar.
2) We need to show two things: a) $U + V$ is a subspace and b) $U + V$ is the least subspace that contains $U \cup V$, or in other words, if $W$ is a subspace of $X$ that contains $U \cup V$, then $U + V \subset W$.
a) Let $\alpha \in K$, $a + b$ and $c + d \in U + V$. We have $\alpha \cdot (a + b) = \alpha \cdot a + \alpha \cdot b \in U + V$ since these are subspaces. Also, $a + b + c + d = (a + c) + (b + d) \in U + V$, once again because they are subspaces. 
b) Let $U + V \subset W$, $a \in U$ and $b \in V$. Then $a = a + 0 \in U + V \subset W$ and $b = 0 + b \in U + V \subset W$ so that $U \cup V \subset W.
Q.E.D.
