Internal direct sum $V = \bigoplus_{i<n} U_i$: the sum is everything and each summand meets the sum of the others only in $0_V$

Let $V$ be a vector space over a field $F$ (Vector space over a field), let $n \in \mathbb{N}$, and let $U$ be a finite family of linear subspaces $U_i$ of $V$ indexed by $i < n$ (Linear subspace of a vector space, The sum $U + W$ of two linear subspaces and the sum $\sum_{i<n} U_i$ of a finite family); as everywhere on this page the index runs over the von Neumann natural $n = \{0, \dots, n-1\}$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$).

The sum of the other summands. The set $\{0_V\}$ is a linear subspace of $V$: it contains $0_V$, it is closed under addition since $0_V + 0_V = 0_V$, and it is closed under scalar multiplication since $\lambda 0_V = 0_V$ (In any vector space $0_F v = 0_V$, $\lambda 0_V = 0_V$, $(-\lambda)v = -(\lambda v)$, $(-1_F)v = -v$, and $\lambda v = 0_V$ forces $\lambda = 0_F$ or $v = 0_V$). So for each $j < n$ the family $U^{(j)}$ defined by

$U^{(j)}_i := U_i \quad (i \ne j), \qquad U^{(j)}_j := \{0_V\}$

is again a finite family of linear subspaces of $V$ indexed by $i < n$, and we write

$\sum_{i \ne j} U_i \;:=\; \sum_{i<n} U^{(j)}_i ,$

a linear subspace of $V$ by The sum $U + W$ of two linear subspaces and the sum $\sum_{i<n} U_i$ of a finite family. Replacing the $j$-th summand by $\{0_V\}$, rather than re-indexing over a smaller set, keeps every family on this page indexed by a natural number.

The definition. $V$ is the internal direct sum of the family $U$, written

$V \;=\; \bigoplus_{i<n} U_i ,$

when both of the following hold:

• (D1) $\displaystyle\sum_{i<n} U_i = V$;
• (D2) for every $j < n$, $\displaystyle U_j \cap \sum_{i \ne j} U_i = \{0_V\}$.

In (D2) the inclusion $\supseteq$ is automatic, since $U_j$ and $\sum_{i \ne j} U_i$ are linear subspaces and each therefore contains $0_V$; the content of (D2) is the inclusion $\subseteq$, that no nonzero vector of $U_j$ is a sum of vectors drawn from the other summands.

Two summands

Take $n = 2$ and write $U := U_0$, $W := U_1$. For $j = 0$ the family $U^{(0)}$ is $\{0_V\}, W$, so $\sum_{i \ne 0} U_i = \{\, 0_V + w : w \in W \,\} = W$; for $j = 1$ it is $U$ in the same way. So (D2) reduces to the single condition $U \cap W = \{0_V\}$, and

$V = U \oplus W \quad\text{means}\quad U + W = V \ \text{ and } \ U \cap W = \{0_V\}.$

For two summands, therefore, (D2) and the pairwise condition coincide; this is the familiar form of the definition.

Three or more summands: (D2) is not the pairwise condition

For $n \ge 3$ the condition (D2) is strictly stronger than requiring $U_i \cap U_j = \{0_V\}$ for all $i \ne j$.

That (D2) implies the pairwise condition is immediate: for $i \ne j$ with $i, j < n$ we have $U_i = U^{(j)}_i \subseteq \sum_{i \ne j} U_i$, since a sum of a family contains each of its summands ($\sum_{i<n} U_i = \operatorname{span}\bigl(\bigcup_{i<n} U_i\bigr)$, so the sum is the smallest linear subspace containing every $U_i$), so $U_j \cap U_i \subseteq U_j \cap \sum_{i \ne j} U_i = \{0_V\}$, and the reverse inclusion holds because both are linear subspaces.

The converse fails, and it fails already for three summands: a family can satisfy (D1) and have all its pairwise intersections trivial while (D2) is false, so that decompositions are not unique. The companion examples page records a witness. A definition stated with the pairwise condition in place of (D2) would therefore be a different, and weaker, notion, and the characterisation by unique decomposition ($V = \bigoplus_{i<n} U_i$ if and only if every $v \in V$ is $\sum_{i<n} u_i$ with $u_i \in U_i$ in exactly one way; equivalently, if and only if the sum is $V$ and $\sum_{i<n} u_i = 0_V$ with $u_i \in U_i$ forces every $u_i = 0_V$) would be false for it.

The empty family

$\mathbb{N}$ contains $0$, so $n = 0$ is a genuine case. Then $\sum_{i<0} U_i = \{0_V\}$ (The sum $U + W$ of two linear subspaces and the sum $\sum_{i<n} U_i$ of a finite family) and (D2) is vacuous, there being no $j < 0$. So $V = \bigoplus_{i<0} U_i$ holds exactly when $V = \{0_V\}$: the zero space is the direct sum of the empty family, and no other space is.

• "Internal" is the operative word. The summands here are linear subspaces of one given space $V$, and the direct sum is a property of that configuration, not a construction producing a new space out of unrelated ones. No external direct sum, and no product of vector spaces, is defined on this page.

• The notation $\bigoplus$ is reserved for the direct case. Writing $\sum_{i<n} U_i$ asserts nothing beyond The sum $U + W$ of two linear subspaces and the sum $\sum_{i<n} U_i$ of a finite family; writing $\bigoplus_{i<n} U_i$ asserts (D1) and (D2) as well. In particular the symbol is not used for a sum that has merely been checked to be everything.

• What (D2) is for. It is exactly the condition that makes decompositions unique: $V = \bigoplus_{i<n} U_i$ holds if and only if every $v \in V$ is $\sum_{i<n} u_i$ with $u_i \in U_i$ in exactly one way. That equivalence is $V = \bigoplus_{i<n} U_i$ if and only if every $v \in V$ is $\sum_{i<n} u_i$ with $u_i \in U_i$ in exactly one way; equivalently, if and only if the sum is $V$ and $\sum_{i<n} u_i = 0_V$ with $u_i \in U_i$ forces every $u_i = 0_V$, and it is the reason the definition is worth stating in this form rather than in terms of uniqueness directly.