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: $m < n \iff m \in n$), so the family is indexed from $0$. Define

$\sum_{i<n} U_i \;:=\; \Bigl\{\, \sum_{i<n} u_i \;:\; u : n \to V \text{ with } u_i \in U_i \text{ for every } i < n \,\Bigr\},$

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 $\operatorname{span}(S)$ 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 $\sum_{i<\sigma(n)} u_i = \bigl(\sum_{i<n} u_i\bigr) + u_n$ (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 $u, w : n \to V$,

$\lambda \sum_{i<n} u_i \;+\; \sum_{i<n} w_i \;=\; \sum_{i<n} (\lambda u_i + w_i).$

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 $\lambda\bigl(\sum_{i<n} u_i + u_n\bigr) + \bigl(\sum_{i<n} w_i + w_n\bigr)$, which by axiom (V2) equals $\bigl(\lambda\sum_{i<n} u_i + \lambda u_n\bigr) + \bigl(\sum_{i<n} w_i + w_n\bigr)$; commutativity and associativity of $+$ regroup this as $\bigl(\lambda\sum_{i<n} u_i + \sum_{i<n} w_i\bigr) + (\lambda u_n + w_n)$, 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 $i \ne j$ and satisfy $u^{(j)}_j = 0_V$. Then

$\sum_{i<n} u_i \;=\; u_j + \sum_{i<n} u^{(j)}_i .$

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: $m < n \iff m \in n$). If $j = n$, then $u^{(n)}$ agrees with $u$ on $n$, so $\sum_{i<\sigma(n)} u^{(n)}_i = \sum_{i<n} u_i + 0_V = \sum_{i<n} u_i$, 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 $\sum_{i<\sigma(n)} u^{(j)}_i = \sum_{i<n} u^{(j)}_i + u_n$, and the inductive hypothesis applied to the restriction of $u$ to $n$ gives $\sum_{i<\sigma(n)} u_i = \bigl(u_j + \sum_{i<n} u^{(j)}_i\bigr) + u_n = u_j + \sum_{i<\sigma(n)} u^{(j)}_i$, by associativity.

A consequence of (F1) and (F3). If $u_i = 0_V$ for every $i \ne j$, 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 $\sum_{i<n} U_i = \operatorname{span}\bigl(\bigcup_{i<n} U_i\bigr)$, so the sum is the smallest linear subspace containing every $U_i$ and of the boundary case of Internal direct sum $V = \bigoplus_{i<n} U_i$: the sum is everything and each summand meets the sum of the others only in $0_V$.

• 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 $V = \bigoplus_{i<n} U_i$ 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 $u_i = 0_V$.

• Why not the union. $\bigcup_{i<n} U_i$ is in general not a linear subspace, and the sum is what repairs that: $\sum_{i<n} U_i = \operatorname{span}\bigl(\bigcup_{i<n} U_i\bigr)$, so the sum is the smallest linear subspace containing every $U_i$ identifies $\sum_{i<n} U_i$ with $\operatorname{span}\bigl(\bigcup_{i<n} U_i\bigr)$, 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 $V = \bigoplus_{i<n} U_i$: 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".