We only call $V=U\oplus W$ a **direct sum** in the special case of a [[sum of subspaces]] that there is a *unique* $u\in V,\ w\in W$ pair that [[vector addition|sum]] to to the [[Element of a Set|element]] $v=u+w$. A sum of subspaces that only intersect at the [[zero vector space]] is a direct sum When taking the **direct sum** of a [[set]] of [[subspace|subspaces]] $V_{1}, \ldots, V_{n}$ of $V$, notation options are $V=\bigoplus_{i=1}^{n} V_{i}=V_{1} \oplus \cdots \oplus V_{n}$ Let $V$ be a [[vector space]] over the [[field]] $K$ Let $U, W$ be subspaces. If $U+W=V$, and if $U \cap W=\{\mathbf0\}$, then $V=U\oplus W$ #### Proof $V$ contains $v=u+w$ a sum of elements from $U$ and $W$ $V \ni v = u+w \quad:\quad u\in U \ \land \ w\in W$ To set up a contradiction, assume that same $v=u^\prime+w^\prime$ a sum of different elements from $U$ and $W$ $V \ni v = u^\prime+w^\prime \quad:\quad u^\prime\in U \ \land \ w^\prime\in W$ Since $v$ is equal to both of these sums, by the transitive property, the sums must be equal $u+w=u^{\prime}+w^{\prime}$ Subtracting $u-u^{\prime}=w^{\prime}-w$ Since $U$ is closed, $(u-u^\prime)\in U$ Since $W$ is closed, $(w-w^\prime)\in W$ And since they are equal, they must be in the intersection, which only contains the [[zero vector]] $(u-u^\prime)=\mathbf0 \implies u=u^\prime$ $(w-w^\prime)=\mathbf0 \implies w=w^\prime$ Which spells doom for more than one distinct sum. Each $v=u+w$ is a unique sum, so $V=U\oplus W$