Theorem

Suppose ${latex.inlineV_{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.inlinev_{1} + ... + v_{m}}, where each ${latex.inlinev_{k} \in V_{k}} is by taking each ${latex.inlinev_{k}} equal to 0.

Proof

First suppose ${latex.inlineV_{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.inlinev_{1} + ... + v_{m}} where each ${latex.inlinev_{k}} equals 0. To show that ${latex.inlineV_{1} + ... + V_{m}} is a direct sum, let ${latex.inlinev \in V_{1} + ... + V_{m}} we can write:

${latex.inlinev = 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.inlinev = 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.inline0 = (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.