Necessary Condition for Direct Sum

Question

Let $V_1, V_2, ..., V_n$ be subspaces of V. How to prove that $V_1 + \cdots + V_n$ is a direct sum if $$\boxed{V_i \cap \sum_{j \neq i} V_j = \{ 0 \} \qquad \textrm{for all $i$}}.$$

Answer 1

Suppose $v_1 + \cdots + v_n = w_1 + \cdots + w_n $ holds for $w_i \in V_i$. We have to show $v_i= w_i$ for all $i = 1, \cdots, n$. Observe that $$ v_i - w_i = \sum_{j \neq i} (w_j - v_j) $$ which says that $v_i - w_i \in V_i \cap \sum_{j \neq i} V_i =\{0\}$. Hence $v_i = w_i$ for all $i= 1, \cdots, n$, and $V_1 + \cdots + V_n$ is a direct sum.