[proofplan] We prove that $U+W$ satisfies the subspace closure conditions inside $V$. The zero vector belongs to $U+W$ because both $U$ and $W$ contain the zero vector. Closure under addition and scalar multiplication follows by writing arbitrary elements of $U+W$ as sums of vectors from $U$ and $W$, then using the corresponding closure properties of $U$ and $W$. Finally, any subspace containing both $U$ and $W$ must contain every vector $u+w$, so it contains $U+W$. [/proofplan] [step:Show that the zero vector belongs to $U+W$] Let $0_V$ denote the zero vector of $V$. Since $U$ and $W$ are subspaces of $V$, both contain $0_V$. Hence $0_V = 0_V + 0_V$ with $0_V \in U$ and $0_V \in W$, so by the definition of $U+W$, \begin{align*} 0_V \in U+W. \end{align*} [/step] [step:Verify closure of $U+W$ under vector addition] Let $x,y \in U+W$. By the definition of $U+W$, there exist $u_1,u_2 \in U$ and $w_1,w_2 \in W$ such that \begin{align*} x = u_1+w_1 \end{align*} and \begin{align*} y = u_2+w_2. \end{align*} Using associativity and commutativity of addition in the [vector space](/page/Vector%20Space) $V$, we obtain \begin{align*} x+y = (u_1+w_1)+(u_2+w_2) = (u_1+u_2)+(w_1+w_2). \end{align*} Since $U$ is a subspace, $u_1+u_2 \in U$. Since $W$ is a subspace, $w_1+w_2 \in W$. Therefore $(u_1+u_2)+(w_1+w_2) \in U+W$, and hence $x+y \in U+W$. [guided] We must prove that adding two arbitrary elements of $U+W$ gives another element of $U+W$. The definition of $U+W$ is the only input needed: each element of $U+W$ is a sum of one vector from $U$ and one vector from $W$. Thus, for $x,y \in U+W$, choose vectors $u_1,u_2 \in U$ and $w_1,w_2 \in W$ satisfying \begin{align*} x = u_1+w_1 \end{align*} and \begin{align*} y = u_2+w_2. \end{align*} Now add these expressions in $V$. The vector space addition on $V$ is associative and commutative, so we may regroup the $U$-terms together and the $W$-terms together: \begin{align*} x+y = (u_1+w_1)+(u_2+w_2) = (u_1+u_2)+(w_1+w_2). \end{align*} This regrouping is useful because the subspace hypotheses apply separately to $U$ and $W$. Since $U$ is closed under addition, $u_1+u_2 \in U$. Since $W$ is closed under addition, $w_1+w_2 \in W$. Therefore the displayed expression writes $x+y$ as a vector from $U$ plus a vector from $W$, so by the definition of $U+W$, \begin{align*} x+y \in U+W. \end{align*} [/guided] [/step] [step:Verify closure of $U+W$ under scalar multiplication] Let $a \in F$ and let $x \in U+W$. By the definition of $U+W$, there exist $u \in U$ and $w \in W$ such that \begin{align*} x = u+w. \end{align*} Using distributivity of scalar multiplication over vector addition in $V$, \begin{align*} a x = a(u+w) = au+aw. \end{align*} Since $U$ is a subspace, $au \in U$. Since $W$ is a subspace, $aw \in W$. Hence $ax \in U+W$. [/step] [step:Conclude that $U+W$ is a subspace of $V$] The set $U+W$ is a subset of $V$ by definition, since every element of $U+W$ has the form $u+w$ with $u,w \in V$. The preceding steps show that $U+W$ contains $0_V$, is closed under addition, and is closed under scalar multiplication by elements of $F$. Therefore $U+W$ is a subspace of $V$. [/step] [step:Prove that $U+W$ is contained in every subspace containing both $U$ and $W$] Let $Z \subset V$ be a subspace such that $U \subset Z$ and $W \subset Z$. Let $x \in U+W$. By the definition of $U+W$, there exist $u \in U$ and $w \in W$ such that \begin{align*} x = u+w. \end{align*} Because $U \subset Z$ and $W \subset Z$, we have $u \in Z$ and $w \in Z$. Since $Z$ is a subspace, it is closed under addition, so $u+w \in Z$. Thus $x \in Z$. Since $x \in U+W$ was arbitrary, \begin{align*} U+W \subset Z. \end{align*} [/step]