#math/linear-algebra Sum of subspaces are defined as $U_{1},U_{2},\dots, U_{m} \subseteq V \text{ where } V \text{ is a vector space.}$. The sum $U_{1} + U_{2} + \dots + U_{m}$ is the set of all possible sums of the elements. $ U_{1} + \dots + U_{m} = \{u_{1} + \dots + u_{m} : u_{1} \in{} U_{1}, \dots, u_{m} \in{} U_{m}\} $ Sums of subspaces are analogous to unions of sets. Unions unions are not used directly with subspaces as they generally do not result in a subspace. - Sums of subspaces, $U_{1} + \dots + U_{m}$, is the smallest subspace which contains each element $U_{1}, \dots, U_{m}$ Direct sums are analogous to disjoint unions of sets. A direct sum is when a sum of subspaces, $U_{1} + \dots + U_{m}$, is when each element can only be written in one way as a sum $u_{1}+\dots + u_{m}$. - Notation that a sum is a direct sum is $U_{1} \oplus \dots \oplus U_{m}$. - $U_{1} + \dots + U_{m}$ is a direct sum if and only if $0$ can only be represented as the sum $u_{1} + \dots + u_{m}$ uniquely (i.e. $u_{1}=\dots=u_{m}=0)$. >[!Unintuitive Trap] > The direct sum of two subspaces has a interesting property in that if $U, W$ are subspaces of $V$. Then $U + W$ is a direct sum if and only if $U \cap W = \{0\}$. However, this idea, possibly unexpectedly, doesn't extend beyond the sum of more than two subspaces and their pairwise intersections (e.g. $U_{1}\cap U_{2} = U_{1}\cap U_{3} = U_{2}\cap U_{3} = \{0\} \nRightarrow U_{1} \oplus U_{2} \oplus U_{3}$).