Problem

Suppose $U_1$ and $U_2$ are subspaces of $V$. Prove that the intersection $U_1 \cap U_2$ is a subspace of $V$.

Solution

  1. $U_1$ and $U_2$ are subspaces and therefore $0 \in U_1$ and $0 \in U_2$ implying that $0 \in U_1 \cap U_2$.
  2. If $x,y \in U_1 \cap U_2$, then $x + y \in U_1$ and $x + y \in U_2$ because these are subspaces. Therefore $x + y \in U_1 \cap U_2$ implying that $U_1 \cap U_2$ is closed under addition.
  3. Similarly if $\lambda \in \mathbb{F}$, then $\lambda x \in U_1$ and $\lambda x \in U_2$ because $U_1$ and $U_2$ are subspaces and $x \in U_1 \cap U_2$, implying that $\lambda x \in U_1 \cap U_2$. Therefore $U_1 \cap U_2$ is closed under scalar multiplication.