Suppose U and W are subspaces of V. Then: \({latex.inline[U + W](U + W)} is a direct sum \){latex.inline\iff} ${latex.inlineU \cap W = {0}}
First suppose U + W is a direct sum. If \({latex.inline[v \in U \cap W](v \in U \cap W)}, then \){latex.inline0 = v + (-v)}, where \({latex.inline[v \in U](v \in U)} and \){latex.inline-v \in W}. By the unique representation of 0 as the sum of a vector in U and a vector in W, we have \({latex.inline[v = 0](v = 0)} . Then we have \){latex.inlineU \cap W = {0}}, completing what we need in one direction.
To go the other way, suppose \({latex.inline[U \cap W = \{0\}](U \cap W = \{0\})}. To prove that \){latex.inlineU + W} is a direct sum, suppose \({latex.inline[u \in U,\ w \in W](u \in U,\ w \in W)} and \){latex.inline0 = u + w}. To complete the proof, we only need to show that \({latex.inline[u = w = 0](u = w = 0)} per [1753142362 - Axler 1.45 Conditions for a direct sum|1.45](1753142362 - Axler 1.45 Conditions for a direct sum|1.45). The equation above implies that \){latex.inlineu = -w \in W}. Thus, \({latex.inline[u \in U \cap W](u \in U \cap W)}. Hence \){latex.inlineu = 0}, which by the equation above implies that ${latex.inlinew = 0}, completing the proof.