Theorem

Suppose U and W are subspaces of V. Then: ${latex.inlineU + W} is a direct sum ${latex.inline\iff} ${latex.inlineU \cap W = {0}}

Proof

First suppose U + W is a direct sum. If ${latex.inlinev \in U \cap W}, then ${latex.inline0 = v + (-v)}, where ${latex.inlinev \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.inlinev = 0} . Then we have ${latex.inlineU \cap W = {0}}, completing what we need in one direction.

To go the other way, suppose ${latex.inlineU \cap W = {0}}. To prove that ${latex.inlineU + W} is a direct sum, suppose ${latex.inlineu \in U,\ w \in W} and ${latex.inline0 = u + w}. To complete the proof, we only need to show that ${latex.inlineu = w = 0} per 1.45. The equation above implies that ${latex.inlineu = -w \in W}. Thus, ${latex.inlineu \in U \cap W}. Hence ${latex.inlineu = 0}, which by the equation above implies that ${latex.inlinew = 0}, completing the proof.