Suppose \({latex.inline[V_{1}, ..., V_{m}](V_{1}, ..., V_{m})} are subspaces of V. Then \){latex.inlineV{1} + ... + V{m}} is a direct sum if and only if the only way to write 0 as a sum \({latex.inline[v_{1} + ... + v_{m}](v_{1} + ... + v_{m})}, where each \){latex.inlinev{k} \in V{k}} is by taking each ${latex.inlinev_{k}} equal to 0.
First suppose \({latex.inline[V_{1} + ... + V_{m}](V_{1} + ... + V_{m})} is a direct sum. Then the definition of direct sum implies that the only way to write 0 and a sum of vectors is by taking each \){latex.inlinev_{k} = 0}.
Now suppose that the only way to write 0 as a sum of \({latex.inline[v_{1} + ... + v_{m}](v_{1} + ... + v_{m})} where each \){latex.inlinev_{k}} equals 0. To show that \({latex.inline[V_{1} + ... + V_{m}](V_{1} + ... + V_{m})} is a direct sum, let \){latex.inlinev \in V{1} + ... + V{m}} we can write:
\({latex.inline[v = v_{1} + ... + v_{m}](v = v_{1} + ... + v_{m})} for some \){latex.inlinev{1} \in V{1}, ..., v{m} \in V{m}}. To show that this representation is unique, suppose we also have \({latex.inline[v = u_{1} + ... + u_{m}](v = u_{1} + ... + u_{m})} where each \){latex.inlineu{1} \in U{1}, ..., u{m} \in U{m}}. Subtracting the two equations, we get that \({latex.inline[0 = (v_{1} - u_{1}) + ... + (v_{m} + u_{m})](0 = (v_{1} - u_{1}) + ... + (v_{m} + u_{m}))}. However, this means that each \){latex.inlinev{i} - u{i} = 0}. Thus, ${latex.inlinev{i} = u{i}} as desired.