The sum $U+W$ of two linear subspaces and the sum ${\sum }_{i<n}{U}_i$ of a finite family

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 of $V$, that is a function assigning to each $i<n$ a linear subspace $U_i$ of $V$ (Linear subspace of a vector space); here $n=\{0,\dots ,n-1\}$ (The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$), so the family is indexed from $0$. Define

${\sum }_{i<n}{U}_i:=\{{\sum }_{i<n}{u}_i:u:n\to V\text{ with }{u}_i\in U_i\text{ for every }i<n\},$

the finite sums being those of The product $g_0{g}_1\cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n=0$) equal to the identity read additively in the abelian group $(V,+,0_V)$, as in Linear combination of a finite list, and the span $\mathrm{span}(S)$ as the smallest linear subspace containing $S$. For two linear subspaces $U,W$ of $V$ we write

$U+W:=\{u+w:u\in U,w\in W\},$

which is the case $n=2$ of the display above, since ${\sum }_{i<2}{u}_i=(0_V+u_0)+u_1=u_0+u_1$.

Three facts about finite sums of vectors

All three are proved by induction on $n$ (The principle of mathematical induction) from the two defining clauses ${\sum }_{i<0}{u}_i=0_V$ and ${\sum }_{i<\sigma (n)}{u}_i=\left({\sum }_{i<n}{u}_i\right)+u_n$ (The product $g_0{g}_1\cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ($n=0$) equal to the identity), together with the abelian group laws of $(V,+,0_V)$. They are collected here because the definition itself needs the first two, and because the lemmas below need all three.

(F1) The all-zero list sums to $0_V$. If $z:n\to V$ has $z_i=0_V$ for every $i<n$, then ${\sum }_{i<n}{z}_i=0_V$. At $n=0$ this is the empty sum, and if it holds at $n$ then ${\sum }_{i<\sigma (n)}{z}_i=0_V+0_V=0_V$.

(F2) The mixed identity. For every $\lambda \in F$ and all lists $u,w:n\to V$,

$\lambda {\sum }_{i<n}{u}_i+{\sum }_{i<n}{w}_i={\sum }_{i<n}(\lambda u_i+w_i)\mathrm{.}$

At $n=0$ both sides are $0_V$, 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 $v=0_V$). If the identity holds at $n$, then at $\sigma (n)$ the left-hand side is $\lambda ({\sum }_{i<n}{u}_i+u_n)+({\sum }_{i<n}{w}_i+w_n)$, which by axiom (V2) equals $(\lambda {\sum }_{i<n}{u}_i+\lambda u_n)+({\sum }_{i<n}{w}_i+w_n)$; commutativity and associativity of $+$ regroup this as $(\lambda {\sum }_{i<n}{u}_i+{\sum }_{i<n}{w}_i)+(\lambda u_n+w_n)$, which by the inductive hypothesis is ${\sum }_{i<n}(\lambda u_i+w_i)+(\lambda u_n+w_n)={\sum }_{i<\sigma (n)}(\lambda u_i+w_i)$.

(F3) Extracting one term. Let $u:n\to V$ and $j<n$, and let $u^{(j)}:n\to V$ agree with $u$ at every $i\mathrm{\ne }j$ and satisfy $u_j^{(j)}=0_V$. Then

${\sum }_{i<n}{u}_i=u_j+{\sum }_{i<n}{u}_i^{(j)}\mathrm{.}$

At $n=0$ there is no $j<0$ and the claim is vacuous. Assume it at $n$ and let $j<\sigma (n)$, so $j\le n$ (On $\mathbb{N}$ the order is membership: $m<n⟺m\in n$). If $j=n$, then $u^{(n)}$ agrees with $u$ on $n$, so ${\sum }_{i<\sigma (n)}{u}_i^{(n)}={\sum }_{i<n}{u}_i+0_V={\sum }_{i<n}{u}_i$, and $u_n+{\sum }_{i<n}{u}_i={\sum }_{i<n}{u}_i+u_n={\sum }_{i<\sigma (n)}{u}_i$ by commutativity. If $j<n$, then $u^{(j)}$ agrees with $u$ at $n$, so ${\sum }_{i<\sigma (n)}{u}_i^{(j)}={\sum }_{i<n}{u}_i^{(j)}+u_n$, and the inductive hypothesis applied to the restriction of $u$ to $n$ gives ${\sum }_{i<\sigma (n)}{u}_i=(u_j+{\sum }_{i<n}{u}_i^{(j)})+u_n=u_j+{\sum }_{i<\sigma (n)}{u}_i^{(j)}$, by associativity.

A consequence of (F1) and (F3). If $u_i=0_V$ for every $i\mathrm{\ne }j$, then $u^{(j)}$ is the all-zero list, so ${\sum }_{i<n}{u}_i=u_j+0_V=u_j$: a list vanishing off a single index sums to its value at that index.

The sum is a linear subspace

${\sum }_{i<n}{U}_i$ is a linear subspace of $V$. It is nonempty: each $U_i$ contains $0_V$, and the all-zero list sums to $0_V$ by (F1), so $0_V\in {\sum }_{i<n}{U}_i$. And it satisfies the one-step test (One-step subspace test: a nonempty $W\subseteq V$ is a linear subspace if and only if $\lambda u+v\in W$ for all $\lambda \in F$ and $u,v\in W$): if $x={\sum }_{i<n}{u}_i$ and $y={\sum }_{i<n}{w}_i$ with $u_i,w_i\in U_i$, and $\lambda \in F$, then (F2) gives $\lambda x+y={\sum }_{i<n}(\lambda u_i+w_i)$, and $\lambda u_i+w_i\in U_i$ because $U_i$ is a linear subspace, so $\lambda x+y\in {\sum }_{i<n}{U}_i$.

So the definition really does produce a linear subspace, and this is asserted here rather than assumed.

The boundary case

$\mathbb{N}$ contains $0$, so $n=0$ is a genuine case. The only list $u:0\to V$ is the empty function, and its sum is the empty sum $0_V$, so

${\sum }_{i<0}{U}_i=\{0_V\},$

the sum of the empty family of linear subspaces being the zero subspace. This is the base case of the induction in ${\sum }_{i<n}{U}_i=\mathrm{span}\left({\bigcup }_{i<n}{U}_i\right)$, so the sum is the smallest linear subspace containing every $U_i$ and of the boundary case of Internal direct sum $V={⨁}_{i<n}{U}_i$: the sum is everything and each summand meets the sum of the others only in $0_V$.

• The sum is a set of vectors, not a set of decompositions. An element of ${\sum }_{i<n}{U}_i$ is a vector that admits at least one expression ${\sum }_{i<n}{u}_i$ with $u_i\in U_i$; different lists may give the same vector, and whether they can is exactly the question answered by $V={⨁}_{i<n}{U}_i$ 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 $u_i=0_V$.

• Why not the union. ${\bigcup }_{i<n}{U}_i$ is in general not a linear subspace, and the sum is what repairs that: ${\sum }_{i<n}{U}_i=\mathrm{span}\left({\bigcup }_{i<n}{U}_i\right)$, so the sum is the smallest linear subspace containing every $U_i$ identifies ${\sum }_{i<n}{U}_i$ with $\mathrm{span}\left({\bigcup }_{i<n}{U}_i\right)$, so the sum is the smallest linear subspace containing every $U_i$.

• (F3) is stated with an index, not with a set. It removes one term from a finite sum by replacing it with $0_V$ rather than by re-indexing the list over a smaller set, which keeps every sum on this page indexed by a von Neumann natural and avoids any appeal to a bijection between index sets. The same device is used in Internal direct sum $V={⨁}_{i<n}{U}_i$: the sum is everything and each summand meets the sum of the others only in $0_V$ to say "the sum of the other summands".