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

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In $F^{3}$ the three coordinate lines are linear subspaces whose internal direct sum is $F^{3}$ , and $F^{0}$ is the zero space

Example

Let $F$ be a field (Field) and consider the vector space $F^{3}$ of functions $3 \to F$ with the pointwise operations (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ). Since $3 = {0, 1, 2}$ (The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ), an element is written $x = (x_{0}, x_{1}, x_{2})$ with $x_{i} : = x (i)$ , indexed from $0$ . For $j < 3$ let $e_{j} \in F^{3}$ be given by $e_{j} (j) = 1_{F}$ and $e_{j} (i) = 0_{F}$ for $i \neq j$ , and put

$L_{j} : = {x \in F^{3} : x_{i} = 0_{F} for every i < 3 with i \neq j} .$

Then:

$L_{j} = {λ e_{j} : λ \in F} = span {e_{j}}$ , so each $L_{j}$ is a linear subspace of $F^{3}$ (Linear subspace of a vector space, Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ );
$F^{3} = ⨁_{j < 3} L_{j}$ (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 unique decomposition of $x \in F^{3}$ being $x = x_{0} e_{0} + x_{1} e_{1} + x_{2} e_{2}$ ;
$F^{0}$ is the zero space: it has exactly one element, the empty function.

The sets $L_{j}$ are called the coordinate lines of $F^{3}$ ; the word "line" is used informally, since dimension is not available here and nothing below uses it.

Facts & Assumptions

Given: A field $F$ , the vector space $F^{3}$ of functions $3 \to F$ with pointwise operations, the vectors $e_{j}$ for $j < 3$ , and the sets $L_{j}$ as displayed.

[L1]

$F^{X}$ is a vector space over $F$ with $(x + y) (i) = x (i) + y (i)$ , $(λ x) (i) = λ x (i)$ and zero the constant function at $0_{F}$ ; for $X = n$ a natural number, $n = {0, \dots, n - 1}$ ; and $F^{0}$ has exactly one element, the empty function, which is its zero vector (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ , Vector space over a field, The natural numbers $N$ (von Neumann), On $N$ the order is membership: $m < n ⟺ m \in n$ ).

[L2]

$span {v} = {λ v : λ \in F}$ , and the span of a subset is a linear subspace ( $span {v} = {λ v : λ \in F}$ , which is ${0_{V}}$ when $v = 0_{V}$ , and when $v \neq 0_{V}$ contains $0_{V}$ only as the multiple $0_{F} v$ , Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ , Linear subspace of a vector space).

[L3]

The elements of $\sum_{i < n} U_{i}$ are exactly the $\sum_{i < n} u_{i}$ with $u_{i} \in U_{i}$ ; and $\sum_{i < 3} u_{i} = (u_{0} + u_{1}) + u_{2}$ , by the recursion $\sum_{i < 0} u_{i} = 0$ and $\sum_{i < σ (m)} u_{i} = (\sum_{i < m} u_{i}) + u_{m}$ together with $0 + u_{0} = u_{0}$ (The sum $U + W$ of two linear subspaces and the sum $\sum_{i < n} U_{i}$ of a finite family).

[L4]

$V = ⨁_{i < n} U_{i}$ 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 ( $V = ⨁_{i < n} U_{i}$ 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 $u_{i} = 0_{V}$ , Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ).

[L5]

In a field: $1_{F} λ = λ$ , multiplication is commutative, $0_{F} λ = 0_{F}$ (Multiplication by zero: $0 \cdot a = 0$ ) and hence $λ 0_{F} = 0_{F}$ , and $0_{F}$ is the additive identity (Field).

Verification

technique · direct

1.1

$F^{3}$ is the set of functions $3 \to F$ with the pointwise operations, and $3 = {0, 1, 2}$ , so the coordinates of an element are $x_{0}, x_{1}, x_{2}$ .

1.2

Each $e_{j}$ is an element of $F^{3}$ , and each $L_{j}$ is a subset of $F^{3}$ , both by their displayed descriptions.

L1L5

1.3

$L_{j} = {λ e_{j} : λ \in F}$ for every $j < 3$ . If $λ \in F$ then $(λ e_{j})_{j} = λ 1_{F} = λ$ and $(λ e_{j})_{i} = λ 0_{F} = 0_{F}$ for $i \neq j$ , so $λ e_{j} \in L_{j}$ . Conversely if $x \in L_{j}$ then $x$ and $x_{j} e_{j}$ have the same value at every $i < 3$ , namely $x_{j}$ at $i = j$ and $0_{F}$ elsewhere, so $x = x_{j} e_{j}$ .

L1L5

1.4

For $u_{0}, u_{1}, u_{2} \in F^{3}$ the finite sum is $\sum_{j < 3} u_{j} = (u_{0} + u_{1}) + u_{2}$ , whose value at $i < 3$ is $(u_{0} (i) + u_{1} (i)) + u_{2} (i)$ , by the pointwise definition of the addition.

L1L3

1.5

$F^{0}$ has exactly one element, the empty function, and that element is its zero vector, so $F^{0}$ is the zero space; this is claim 3.

2.1

Each $L_{j}$ is a linear subspace of $F^{3}$ and equals $span {e_{j}}$ : by step 1.3 it is the set of scalar multiples of $e_{j}$ , which is exactly the span of ${e_{j}}$ , and a span is a linear subspace. This is claim 1.

step 1.3L2

2.2

Every $x \in F^{3}$ decomposes. Put $u_{j} : = x_{j} e_{j}$ , which lies in $L_{j}$ by step 1.3. The value of $\sum_{j < 3} u_{j}$ at $i < 3$ is $(x_{0} e_{0} (i) + x_{1} e_{1} (i)) + x_{2} e_{2} (i)$ ; since $e_{j} (i) = 0_{F}$ for $j \neq i$ and $e_{i} (i) = 1_{F}$ , exactly one summand is $x_{i}$ and the others are $0_{F}$ , so the value is $x_{i}$ . Hence $\sum_{j < 3} u_{j} = x$ , and $\sum_{j < 3} L_{j} = F^{3}$ .

step 1.3step 1.4L1L5

3.1

The decomposition is unique. Suppose $u_{j} \in L_{j}$ for $j < 3$ and $\sum_{j < 3} u_{j} = x$ . Evaluating at $i < 3$ gives $(u_{0} (i) + u_{1} (i)) + u_{2} (i) = x_{i}$ , and $u_{j} (i) = 0_{F}$ whenever $j \neq i$ , so the left-hand side is $u_{i} (i)$ ; thus $u_{i} (i) = x_{i}$ , and step 1.3 gives $u_{i} = u_{i} (i) e_{i} = x_{i} e_{i}$ . So the list is the one of step 2.2.

step 1.3step 1.4L1L5

4.1

By steps 2.2 and 3.1 every $x \in F^{3}$ is $\sum_{j < 3} u_{j}$ with $u_{j} \in L_{j}$ in exactly one way, so $F^{3} = ⨁_{j < 3} L_{j}$ , and the decomposition is $x = x_{0} e_{0} + x_{1} e_{1} + x_{2} e_{2}$ . This is claim 2.

step 2.2step 3.1L4

5.1

Claim 1 is step 2.1, claim 2 is step 4.1 and claim 3 is step 1.5.

step 1.5step 2.1step 4.1∎

Remarks

Every index here starts at $0$ . The coordinates are $x_{0}, x_{1}, x_{2}$ and the summands are $L_{0}, L_{1}, L_{2}$ , because $3 = {0, 1, 2}$ as a von Neumann natural. Writing the same example with coordinates $x_{1}, x_{2}, x_{3}$ would not match the definition of $F^{n}$ used here (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ ).
$F^{0}$ is not empty. There is exactly one function $\emptyset \to F$ , so $F^{0}$ has exactly one element and is the zero space. It is also the internal direct sum of the empty family of its linear subspaces (Internal direct sum $V = ⨁_{i < n} U_{i}$ : the sum is everything and each summand meets the sum of the others only in $0_{V}$ ), which is the only space that is.
Nothing above is special to $3$ . The same computation works for any $n \in N$ and gives $F^{n} = ⨁_{j < n} L_{j}$ , at $n = 0$ degenerating to the statement that the zero space is the direct sum of the empty family. It is written out at $n = 3$ so that the finite sums are the explicit $(u_{0} + u_{1}) + u_{2}$ rather than an induction.

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Sources

Examples of vector spaces (Wikipedia) (standard reference, not scraped)
Direct sum of modules (Wikipedia) (standard reference, not scraped)
S. Axler, Linear Algebra Done Right, 4th ed. (free PDF, CC BY-NC) (standard reference, not scraped)

In F3F3 the three coordinate lines are linear subspaces whose internal direct sum is F3F3, and F0F0 is the zero space