[proofplan] Both parts are verified directly from the subspace axioms (non-emptiness, [closure](/page/Closure) under addition and scalar multiplication). For (i), membership in both $U$ and $W$ is preserved by the operations since each subspace is individually closed. For (ii), sums and scalar multiples of elements $u + w$ decompose into components from $U$ and $W$. Minimality follows because any subspace containing both $U$ and $W$ is closed under addition. [/proofplan] [step:Verify the subspace axioms for $U \cap W$] Since $U$ and $W$ are subspaces, $\mathbf{0} \in U$ and $\mathbf{0} \in W$, so $\mathbf{0} \in U \cap W$ and the intersection is non-empty. Let $v_1, v_2 \in U \cap W$ and $\lambda \in \mathbb{F}$. Since $v_1, v_2 \in U$ and $U$ is a subspace: $v_1 + v_2 \in U$ and $\lambdav_1 \in U$. Since $v_1, v_2 \in W$ and $W$ is a subspace: $v_1 + v_2 \in W$ and $\lambdav_1 \in W$. Hence $v_1 + v_2 \in U \cap W$ and $\lambdav_1 \in U \cap W$. [/step] [step:Verify the subspace axioms for $U + W$] Since $\mathbf{0} \in U$ and $\mathbf{0} \in W$, we have $\mathbf{0} = \mathbf{0} + \mathbf{0} \in U + W$. Let $v_1 = u_1 + w_1$ and $v_2 = u_2 + w_2$ be elements of $U + W$ with $u_1, u_2 \in U$ and $w_1, w_2 \in W$. Then: \begin{align*} v_1 + v_2 = (u_1 + u_2) + (w_1 + w_2) \in U + W, \end{align*} since $u_1 + u_2 \in U$ (closure of $U$) and $w_1 + w_2 \in W$ (closure of $W$). For scalar multiplication with $\lambda \in \mathbb{F}$: \begin{align*} \lambdav_1 = \lambdau_1 + \lambdaw_1 \in U + W, \end{align*} since $\lambdau_1 \in U$ and $\lambdaw_1 \in W$. [/step] [step:Prove $U + W$ is the smallest subspace containing both $U$ and $W$] First, $U \subset U + W$ (write $u = u + \mathbf{0}$) and $W \subset U + W$ (write $w = \mathbf{0} + w$), so $U + W$ contains both $U$ and $W$. Let $X \subset V$ be any subspace with $U \subset X$ and $W \subset X$. For any $u + w \in U + W$, we have $u \in U \subset X$ and $w \in W \subset X$, so $u + w \in X$ by closure of $X$ under addition. Hence $U + W \subset X$, establishing minimality. [/step]