A set $S$ is the ${disjoint}$ sum of sets $A$ and $B$ if the following is true.
- $S = A \cup B$ and
- $A$ and $B$ have no elements in common, i.e. $A \cap B = \emptyset$.
Now let $V$ be a vector-space over a field $\mathbb{F}$. We want to say that $V$ is the 'disjoint' sum of two subspaces $U, W$ of $V$. We cannot say that $U$ and $W$ are disjoint as sets because any subspace of $V$ contains $0$. What we can do is to minimise their intersection: other than the 0-vector, $U$ and $W$ should not have any common members. Moreover, to get all of $V$, we must require that every vector $v$ in $V$ can somehow be constructed from elements of $U$ and $W$. This gives rise to the following definition of direct sum:
- $S = U + W$ and
- $U$ and $W$ have no non-zero elements in common, i.e. $U \cap W = \{0\}$.
Clearly this definition of direct sum in a vector space is almost the same as that of disjoint sum of sets. The only thing that remains in need of explanation is the sum of subspaces $U + W$. But that's simple:
$U + W = \{u + w\ |\ u \in U, w \in W\}$
In other words, we are lifting vector addition from vectors to sets of vectors.
As an example, the vector space $\mathbb{R}^2$, i.e. the euclidean plane, is the direct sum of the x-axis and the y-axis. In other words $\mathbb{R}^2$ is the direct sum of the subspace $\{ (x, 0)\ |\ x \in \mathbb{R}\}$ and $\{ (0, y)\ |\ y \in \mathbb{R}\}$.
It turns out that the formal similarity between these two constructions is not accidental. Both are instances of a more general construction called co-product or categorical sum which is defined for categories.