LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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$V = ⨁_{i < n} U_{i}$ 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 $u_{i} = 0_{V}$

Statement

Let $V$ be a vector space over a field $F$ (Vector space over a field), let $n \in N$ , and let $U$ be a finite family of linear subspaces $U_{i}$ of $V$ indexed by $i < n$ (The sum $U + W$ of two linear subspaces and the sum $\sum_{i < n} U_{i}$ of a finite family). Call a list $u : n \to V$ admissible when $u_{i} \in U_{i}$ for every $i < n$ . The following are equivalent.

(a) $V = ⨁_{i < n} U_{i}$ (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).
(b) For every $v \in V$ there is exactly one admissible list $u$ with $\sum_{i < n} u_{i} = v$ .
(c) $\sum_{i < n} U_{i} = V$ , and the only admissible list $u$ with $\sum_{i < n} u_{i} = 0_{V}$ is the list with $u_{i} = 0_{V}$ for every $i < n$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ , a natural number $n$ , and a finite family of linear subspaces $U_{i}$ of $V$ indexed by $i < n$ ; a list $u : n \to V$ is called admissible when $u_{i} \in U_{i}$ for every $i < n$ .

[L1]

$V = ⨁_{i < n} U_{i}$ means (D1) $\sum_{i < n} U_{i} = V$ and (D2) $U_{j} \cap \sum_{i \neq j} U_{i} = {0_{V}}$ for every $j < n$ , where $\sum_{i \neq j} U_{i} = \sum_{i < n} U_{i}^{(j)}$ for the family $U^{(j)}$ with $U_{i}^{(j)} = U_{i}$ for $i \neq j$ and $U_{j}^{(j)} = {0_{V}}$ (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).

[L2]

The elements of $\sum_{i < n} U_{i}$ are exactly the vectors $\sum_{i < n} u_{i}$ with $u$ admissible; it is a linear subspace of $V$ ; the mixed identity (F2) $λ \sum_{i < n} u_{i} + \sum_{i < n} w_{i} = \sum_{i < n} (λ u_{i} + w_{i})$ holds; and by (F3) with (F1), $\sum_{i < n} u_{i} = u_{j} + \sum_{i < n} u_{i}^{(j)}$ for $j < n$ , where $u^{(j)}$ agrees with $u$ off $j$ and has $u_{j}^{(j)} = 0_{V}$ , while a list vanishing off a single index sums to its value there (The sum $U + W$ of two linear subspaces and the sum $\sum_{i < n} U_{i}$ of a finite family, The product $g_{0} g_{1} \dots g_{n - 1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

[L3]

A linear subspace contains $0_{V}$ and is closed under $+$ and under scalar multiplication (Linear subspace of a vector space).

[L4]

$(- 1_{F}) x = - x$ for every $x \in V$ , and $1_{F} x = x$ (In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ , Vector space over a field).

[L5]

Cancellation in the abelian group $(V, +, 0_{V})$ : if $x + y = x + z$ then $y = z$ , and if $y + x = z + x$ then $y = z$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution, Group and abelian group).

[L6]

$(V, +, 0_{V})$ is an abelian group: $+$ is associative and commutative, $0_{V}$ is a two-sided identity, and each $x$ has an additive inverse $- x$ with $x + (- x) = 0_{V} = (- x) + x$ (Vector space over a field, Group and abelian group).

[L7]

The index $i$ runs over the von Neumann natural $n = {0, \dots, n - 1}$ (The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ).

Proof

technique · direct

1.1

Let $u$ be admissible and $j < n$ . The list $u^{(j)}$ is admissible for the family $U^{(j)}$ , since $u_{i}^{(j)} = u_{i} \in U_{i} = U_{i}^{(j)}$ for $i \neq j$ and $u_{j}^{(j)} = 0_{V} \in {0_{V}} = U_{j}^{(j)}$ ; hence $s : = \sum_{i < n} u_{i}^{(j)}$ lies in $\sum_{i \neq j} U_{i}$ , and $\sum_{i < n} u_{i} = u_{j} + s$ .

L1L2L7

1.2

A linear subspace $W$ of $V$ is closed under additive inverses: for $x \in W$ we have $(- 1_{F}) x \in W$ by closure under scalar multiplication, and $(- 1_{F}) x = - x$ .

L3L4

1.3

Let $j < n$ and $x \in U_{j}$ . The list $z : n \to V$ with $z_{j} = x$ and $z_{i} = 0_{V}$ for $i \neq j$ is admissible, each $U_{i}$ containing $0_{V}$ , and it sums to $x$ .

L2L3L7

1.4

(c) implies (b). Assume (c). Existence: since $\sum_{i < n} U_{i} = V$ , every $v \in V$ is $\sum_{i < n} u_{i}$ for some admissible $u$ . Uniqueness: suppose $u$ and $w$ are admissible with $\sum_{i < n} u_{i} = \sum_{i < n} w_{i} = : v$ . The mixed identity with $λ = - 1_{F}$ , applied to $w$ and $u$ in that order, gives $(- 1_{F}) \sum_{i < n} w_{i} + \sum_{i < n} u_{i} = \sum_{i < n} ((- 1_{F}) w_{i} + u_{i})$ , whose left-hand side is $- v + v = 0_{V}$ ; the list $i \mapsto (- 1_{F}) w_{i} + u_{i} = (- w_{i}) + u_{i}$ is admissible, each $U_{i}$ being closed under additive inverses and addition; so by (c) every $(- w_{i}) + u_{i} = 0_{V} = (- w_{i}) + w_{i}$ , and cancelling $- w_{i}$ on the left gives $u_{i} = w_{i}$ for every $i < n$ .

L2L3L4L5L6

2.1

(a) implies (c). Assume (a). Condition (D1) is the first half of (c). For the second, let $u$ be admissible with $\sum_{i < n} u_{i} = 0_{V}$ and let $j < n$ . Writing $s = \sum_{i < n} u_{i}^{(j)} \in \sum_{i \neq j} U_{i}$ , we get $u_{j} + s = 0_{V}$ , while $(- s) + s = 0_{V}$ as well, so cancelling $s$ on the right gives $u_{j} = - s$ ; and $- s \in \sum_{i \neq j} U_{i}$ because that set is a linear subspace. Hence $u_{j} \in U_{j} \cap \sum_{i \neq j} U_{i}$ , which is ${0_{V}}$ by (D2), so $u_{j} = 0_{V}$ . As $j < n$ was arbitrary, $u$ is the all-zero list.

step 1.1step 1.2L1L2L5L6

2.2

(b) implies (a). Assume (b). For (D1): every $v \in V$ is $\sum_{i < n} u_{i}$ for some admissible $u$ , so $V \subseteq \sum_{i < n} U_{i}$ , and the reverse inclusion holds because $\sum_{i < n} U_{i}$ is a subset of $V$ . For (D2): let $j < n$ and $x \in U_{j} \cap \sum_{i \neq j} U_{i}$ . Then $x = \sum_{i < n} w_{i}$ for some list $w$ admissible for $U^{(j)}$ ; such a $w$ has $w_{j} = 0_{V}$ and $w_{i} \in U_{i}$ for $i \neq j$ , so it is admissible for $U$ as well, $U_{j}$ containing $0_{V}$ . The list $z$ of step 1.3 is also admissible and also sums to $x$ , so uniqueness in (b) forces $z = w$ , and in particular $x = z_{j} = w_{j} = 0_{V}$ . Since ${0_{V}}$ is contained in the intersection anyway, (D2) holds.

step 1.3L1L2L3

3.1

Steps 2.1, 1.4 and 2.2 give (a) implies (c), (c) implies (b) and (b) implies (a), so the three conditions are equivalent.

step 1.4step 2.1step 2.2∎

Remarks

Condition (c) is the one used in practice. Checking uniqueness of every decomposition is checking a single one: that of $0_{V}$ . The reduction is the content of the implication from (c) to (b), and it works because the difference of two admissible decompositions of the same vector is an admissible decomposition of $0_{V}$ .
This is what makes (D2) the right condition. If the definition of a direct sum had asked only for pairwise trivial intersections, the equivalence above would fail for $n \geq 3$ : the companion examples page exhibits three linear subspaces of a plane whose pairwise intersections are trivial, whose sum is everything, and for which some vector has two different decompositions. So the equivalence proved here is not available for the pairwise notion, and (D2) is exactly the strengthening that restores it.
The two-summand case reads as usual. For $n = 2$ , condition (a) says $U + W = V$ and $U \cap W = {0_{V}}$ (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ), and the lemma says that this holds exactly when every $v \in V$ is $u + w$ with $u \in U$ and $w \in W$ in exactly one way.
No finiteness of $V$ and no dimension anywhere. The family of summands is finite because the sum $\sum_{i < n} U_{i}$ is defined through a finite sum of vectors; $V$ itself is arbitrary, and nothing above counts anything.

Depends on

Used by

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Direct sum of modules (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed., Ch. 1 (standard reference, not scraped)
Direct sum (Encyclopedia of Mathematics) (standard reference, not scraped)

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