1. **State the problem:** Prove that the sum $W_1 + W_2$ of two subspaces $W_1$ and $W_2$ is a direct sum if and only if their intersection is the zero vector space, i.e., $W_1 \cap W_2 = \{0\}$. 2. **Recall definitions:** - The sum $W_1 + W_2$ is defined as $\{w_1 + w_2 : w_1 \in W_1, w_2 \in W_2\}$. - The sum is a direct sum, denoted $W_1 \oplus W_2$, if every element in $W_1 + W_2$ can be written uniquely as $w_1 + w_2$. 3. **To prove:** $W_1 + W_2$ is a direct sum $\iff W_1 \cap W_2 = \{0\}$. 4. **Proof:** - ($\Rightarrow$) Assume $W_1 + W_2$ is a direct sum. Suppose $v \in W_1 \cap W_2$. Then $v \in W_1$ and $v \in W_2$. Consider the zero vector $0 \in W_1 + W_2$. It can be written as $0 = v + (-v)$ with $v \in W_1$ and $-v \in W_2$. But also $0 = 0 + 0$ with $0 \in W_1$ and $0 \in W_2$. Since the sum is direct, the representation is unique, so $v = 0$. Hence $W_1 \cap W_2 = \{0\}$. - ($\Leftarrow$) Assume $W_1 \cap W_2 = \{0\}$. Suppose $x \in W_1 + W_2$ has two representations: $x = w_1 + w_2 = w_1' + w_2'$ with $w_1, w_1' \in W_1$ and $w_2, w_2' \in W_2$. Then $$w_1 - w_1' = w_2' - w_2.$$ The left side is in $W_1$ and the right side is in $W_2$. Since $W_1 \cap W_2 = \{0\}$, this implies $$w_1 - w_1' = 0 \quad \Rightarrow \quad w_1 = w_1'$$ and $$w_2' - w_2 = 0 \quad \Rightarrow \quad w_2 = w_2'.$$ Thus, the representation is unique and $W_1 + W_2$ is a direct sum. 5. **Conclusion:** We have shown both directions, so $W_1 + W_2$ is a direct sum if and only if $W_1 \cap W_2 = \{0\}$.