The definition of direct sum requires every vector in the sum to have a unique representation as an appropriate sum. The next result shows that when deciding whether a sum of subspaces is a direct sum, we only need to consider whether can be uniquely written as an appropriate sum.

Condition for Direct Sums

Suppose are subspaces of . Then, is a direct sum if and only if the only way to write as a sum , where each , is by taking each equal to .

Proof.

First, we show necessity: assuming that is a direct sum, the only unique way to write the zero vector is if each .

Suppose is a direct sum. All of the subspaces are subspaces of , which means all of these subspaces share the same additive identity . Clearly, adding all of these additive identities , where each , is equal to . Then, the definition of direct sum implies that this is the only way to write , since this must unique.

Thus, the definition of direct sum implies that the only way to write as a sum , where each , is by taking each equal to .

Second, we show sufficiency: assuming that the only way to express the zero vectors is by having all , this implies that is a direct sum.

Suppose that the only way to write as a sum , where , is by taking each equal to .

To show that is a direct sum, let some . We can write

for some . To show that this representation is unique, suppose we also have

where . Subtracting these two equations, we have:

Because , the equation above implies that each equals . This shows that any two representations of the same vector must be equal, so every vector actually only has one unique representation.