ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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The vector $(1,2) \in \mathbb{R}^{2}$ has coordinate list $(1,2)$ in the standard ordered basis, $(2,1)$ in its reversal, and $(2,-1)$ in the ordered basis $\bigl((1,1),(1,0)\bigr)$

Example

Let $\mathbb{R}$ be the real numbers (The real numbers), a field (The reals form a field), and let $\mathbb{R}^{2}$ be the function space on the von Neumann natural $2 = \{0,1\}$ with the pointwise operations (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = \{0, 1, \dots, n-1\}$ , The natural numbers $\mathbb{N}$ (von Neumann), On $\mathbb{N}$ the order is membership: $m < n \iff m \in n$ ). We write $(y_0, y_1)$ for the element $y$ of $\mathbb{R}^{2}$ with $y(0) = y_0$ and $y(1) = y_1$ , so that $e_0 = (1,0)$ and $e_1 = (0,1)$ are the standard unit vectors (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ ).

Put $x := (1,2)$ and consider three ordered bases of $\mathbb{R}^{2}$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis):

$e = (e_0, e_1)$ , the standard ordered basis;
$e' = (e_1, e_0)$ , its reversal, which has the same image $\{e_0, e_1\}$ ;
$v = (v_0, v_1)$ with $v_0 := (1,1)$ and $v_1 := (1,0)$ .

Then the coordinate list of $x$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i<n} \lambda_i v_i$ for exactly one $\lambda : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis) is

$(1,2) \ \text{ in } e, \qquad (2,1) \ \text{ in } e', \qquad (2,-1) \ \text{ in } v .$

Three different lists for one vector, and the first two differ although the two ordered bases have the same image. Coordinates are attached to an ordered basis, not to a basis.

Facts & Assumptions

Given: The field $\mathbb{R}$ , the vector space $\mathbb{R}^{2}$ with pointwise operations, the vector $x = (1,2)$ , and the three lists $e$ , $e'$ and $v$ above.

[L1]

$F^{X}$ is a vector space over $F$ with pointwise operations, and two elements are equal exactly when they agree at every point (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).

[L2]

$e : 2 \to \mathbb{R}^{2}$ is an ordered basis, and $\bigl(\sum_{i<2}\lambda_i e_i\bigr)(j) = \lambda_j$ for every $\lambda : 2 \to \mathbb{R}$ and $j < 2$ (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ , claims 2 and 3).

[L3]

A list $w : n \to V$ is an ordered basis if and only if every $y \in V$ is $\sum_{i<n}\lambda_i w_i$ for exactly one $\lambda : n \to F$ , and that $\lambda$ is then the coordinate list of $y$ (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i<n} \lambda_i v_i$ for exactly one $\lambda : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis, Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L4]

Finite sums: $\sum_{i<2}u_i = (0_V + u_0) + u_1 = u_0 + u_1$ (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, Linear combination of a finite list, and the span $\operatorname{span}(S)$ as the smallest linear subspace containing $S$ ).

[L5]

The vector space axioms and the field axioms of $\mathbb{R}$ : (V2) $\lambda(y+z) = \lambda y + \lambda z$ , (V3) $(\lambda+\mu)y = \lambda y + \mu y$ , (V5) $1y = y$ ; $(V,+,0_V)$ is abelian; $0y = 0_V$ ; and $\mathbb{R}$ is a field (Vector space over a field, 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 $v = 0_V$ , Field, The reals form a field).

[L6]

Injectivity and images are as in Injection, surjection, bijection.

Verification

technique · direct

1.1

Coordinates in $e$ . By the standard basis lemma, $e$ is an ordered basis of $\mathbb{R}^{2}$ and the coordinate list of $y \in \mathbb{R}^{2}$ is $i \mapsto y(i)$ ; for $x = (1,2)$ that list is $(1,2)$ .

L2L3

1.2

$e'$ is an ordered basis. The list $e' = (e_1, e_0)$ is injective, since $e_0 \ne e_1$ ( $e_0$ takes the value $1$ at $0$ and $e_1$ takes the value $0$ there), and its image is $\{e_0, e_1\} = e[2]$ , which is a basis of $\mathbb{R}^{2}$ ; so $e'$ is an injective list whose image is a basis.

L1L2L6

1.3

Coordinates in $e'$ . For $\lambda : 2 \to \mathbb{R}$ , $\sum_{i<2}\lambda_i e'_i = \lambda_0 e_1 + \lambda_1 e_0 = \lambda_1 e_0 + \lambda_0 e_1 = \sum_{i<2}\mu_i e_i$ with $\mu_0 = \lambda_1$ and $\mu_1 = \lambda_0$ ; evaluating with the standard basis, this vector is $(\lambda_1, \lambda_0)$ . It equals $x = (1,2)$ exactly when $\lambda_1 = 1$ and $\lambda_0 = 2$ , so the coordinate list of $x$ in $e'$ is $(2,1)$ .

L2L4L5

1.4

$v$ is an ordered basis and the coordinates of a general vector in it. Note $v_0 = (1,1) = e_0 + e_1$ and $v_1 = (1,0) = e_0$ . For $\lambda : 2 \to \mathbb{R}$ , $\sum_{i<2}\lambda_i v_i = \lambda_0(e_0+e_1) + \lambda_1 e_0 = (\lambda_0 e_0 + \lambda_0 e_1) + \lambda_1 e_0 = (\lambda_0 + \lambda_1)e_0 + \lambda_0 e_1$ , using (V2), (V3) and the abelian group laws; by the standard basis this vector is $(\lambda_0 + \lambda_1,\ \lambda_0)$ . Given $y = (y_0, y_1)$ , the equations $\lambda_0 + \lambda_1 = y_0$ and $\lambda_0 = y_1$ have the unique solution $\lambda_0 = y_1$ , $\lambda_1 = y_0 - y_1$ , so every $y$ is $\sum_{i<2}\lambda_i v_i$ for exactly one $\lambda$ and $v$ is an ordered basis.

L2L3L4L5

2.1

Coordinates of $x$ in $v$ . Taking $y = x = (1,2)$ in step 1.4 gives $\lambda_0 = 2$ and $\lambda_1 = 1 - 2 = -1$ , so the coordinate list of $x$ in $v$ is $(2,-1)$ ; and $2(1,1) + (-1)(1,0) = (2,2) + (-1,0) = (1,2) = x$ confirms it.

step 1.4L1L5

3.1

The three coordinate lists of the single vector $x$ are therefore $(1,2)$ , $(2,1)$ and $(2,-1)$ , computed in steps 1.1, 1.3 and 2.1; the first two are different although $e$ and $e'$ have the same image, so the coordinate list depends on the ordered basis and not merely on the underlying set.

step 1.1step 1.3step 2.1∎

Remarks

What is and is not being said. Uniqueness of the coordinate list (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i<n} \lambda_i v_i$ for exactly one $\lambda : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis) is uniqueness for a fixed ordered basis. Nothing there says that different ordered bases give the same list, and this example shows they do not, even when they differ only in the order. Reordering the list permutes the coordinates of every vector at once.
The third basis is not a reordering of the first. Its image $\{(1,1),(1,0)\}$ is a different set from $\{(1,0),(0,1)\}$ , and its coordinates differ for a further reason: the vectors themselves are different. The passage between coordinate lists of two ordered bases is a change of basis, taken up on a later page once linear maps are available; the point here is only that the two lists differ.
The arithmetic was recomputed, not copied. With $v_0 = (1,1)$ and $v_1 = (1,0)$ , matching $(1,2)$ forces the second coordinate first: $\lambda_0$ is the second entry, so $\lambda_0 = 2$ , and then $\lambda_1 = 1 - 2 = -1$ . Reading the pair off in the other order would give $(-1,2)$ , which is wrong.

Depends on

Used by

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Sources

Basis (linear algebra) (Wikipedia) (standard reference, not scraped)
Coordinate vector (Wikipedia) (standard reference, not scraped)

The vector (1,2)∈R2(1,2) \in \mathbb{R}^{2}(1,2)∈R2 has coordinate list (1,2)(1,2)(1,2) in the standard ordered basis, (2,1)(2,1)(2,1) in its reversal, and (2,−1)(2,-1)(2,−1) in the ordered basis ((1,1),(1,0))\bigl((1,1),(1,0)\bigr)((1,1),(1,0))