1. **Problem Statement:** Prove that if $W_1$ and $W_2$ are subspaces of a vector space $V$, then their intersection $W_1 \cap W_2$ is also a subspace of $V$. 2. **Recall the definition of a subspace:** A subset $U$ of a vector space $V$ is a subspace if it satisfies three conditions: - The zero vector $\mathbf{0}$ of $V$ is in $U$. - $U$ is closed under vector addition: if $\mathbf{u}, \mathbf{v} \in U$, then $\mathbf{u} + \mathbf{v} \in U$. - $U$ is closed under scalar multiplication: if $\mathbf{u} \in U$ and $c$ is a scalar, then $c\mathbf{u} \in U$. 3. **Show $W_1 \cap W_2$ contains the zero vector:** Since $W_1$ and $W_2$ are subspaces, both contain the zero vector $\mathbf{0}$. Therefore, $\mathbf{0} \in W_1$ and $\mathbf{0} \in W_2$, so $\mathbf{0} \in W_1 \cap W_2$. 4. **Show closure under addition:** Let $\mathbf{u}, \mathbf{v} \in W_1 \cap W_2$. By definition of intersection, $\mathbf{u}, \mathbf{v} \in W_1$ and $\mathbf{u}, \mathbf{v} \in W_2$. Since $W_1$ is a subspace, $\mathbf{u} + \mathbf{v} \in W_1$. Since $W_2$ is a subspace, $\mathbf{u} + \mathbf{v} \in W_2$. Therefore, $\mathbf{u} + \mathbf{v} \in W_1 \cap W_2$. 5. **Show closure under scalar multiplication:** Let $\mathbf{u} \in W_1 \cap W_2$ and $c$ be any scalar. Since $\mathbf{u} \in W_1$ and $W_1$ is a subspace, $c\mathbf{u} \in W_1$. Since $\mathbf{u} \in W_2$ and $W_2$ is a subspace, $c\mathbf{u} \in W_2$. Therefore, $c\mathbf{u} \in W_1 \cap W_2$. 6. **Conclusion:** Since $W_1 \cap W_2$ contains the zero vector and is closed under addition and scalar multiplication, it is a subspace of $V$. **Final answer:** The intersection $W_1 \cap W_2$ of two subspaces $W_1$ and $W_2$ of a vector space $V$ is itself a subspace of $V$.