Internal direct sum $V={⨁}_{i<n}{U}_i$: 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: $m<n⟺m\in n$).

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

$U_i^{(j)}:=U_i(i\mathrm{\ne }j),U_j^{(j)}:=\{0_V\}$

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

${\sum }_{i\mathrm{\ne }j}{U}_i:={\sum }_{i<n}{U}_i^{(j)},$

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

$V={⨁}_{i<n}{U}_i,$

when both of the following hold:

• (D1) ${\displaystyle \sum _{i<n}{U}_i=V}$;
• (D2) for every $j<n$, ${\displaystyle U_j\cap \sum _{i\mathrm{\ne }j}{U}_i=\{0_V\}}$.

In (D2) the inclusion $\supseteq$ is automatic, since $U_j$ and ${\sum }_{i\mathrm{\ne }j}{U}_i$ 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 ${\sum }_{i\mathrm{\ne }0}{U}_i=\{0_V+w:w\in W\}=W$; for $j=1$ it is $U$ in the same way. So (D2) reduces to the single condition $U\cap W=\{0_V\}$, and

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

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 $i\mathrm{\ne }j$.

That (D2) implies the pairwise condition is immediate: for $i\mathrm{\ne }j$ with $i,j<n$ we have $U_i=U_i^{(j)}\subseteq {\sum }_{i\mathrm{\ne }j}{U}_i$, since a sum of a family contains each of its summands (${\sum }_{i<n}{U}_i=\mathrm{span}\left({\bigcup }_{i<n}{U}_i\right)$, so the sum is the smallest linear subspace containing every $U_i$), so $U_j\cap U_i\subseteq U_j\cap {\sum }_{i\mathrm{\ne }j}{U}_i=\{0_V\}$, 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 ($V={⨁}_{i<n}{U}_i$ 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 $V={⨁}_{i<0}{U}_i$ 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 $⨁$ 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 ${⨁}_{i<n}{U}_i$ 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: $V={⨁}_{i<n}{U}_i$ 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 $V={⨁}_{i<n}{U}_i$ 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.