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Mathematics · 术语表

什么是 linear form?

又称: hyperplane

Definition 20.17 University Mathematics — Year 1 · 第 20 章 — Linear Maps

A linear form on EE is a linear map φ ⁣:E→K\varphi \colon E \to K. A hyperplane of EE (dim⁡E=n\dim E = n) is a subspace of dimension n−1n - 1.

示例

Example 20.18 (An evaluation form and its hyperplane)

On R2[X]\R_2[X], the evaluation φ(P)=P(2)\varphi(P) = P(2) is a linear form, nonzero (φ(1)=1\varphi(1) = 1). Its kernel is the hyperplane of polynomials vanishing at 22, i.e. (factor theorem, Theorem 8.7) the multiples of X−2X - 2 within R2[X]\R_2[X]:

ker⁡φ=Vect⁡(X−2, X(X−2)),dim⁡=2.\ker\varphi = \operatorname{Vect}\bigl(X - 2,\ X(X - 2)\bigr), \qquad \dim = 2 .

In coordinates on (1,X,X2)(1, X, X^2), φ(a+bX+cX2)=a+2b+4c\varphi(a + bX + cX^2) = a + 2b + 4c: every linear form on a finite-dimensional space is, once a basis is fixed, a fixed linear expression in the coordinates — forms are “row vectors”, as Chapter 21 will make literal, and the coefficient row here, (1,2,4)(1, 2, 4), is a Vandermonde row: evaluation forms are how the interpolation theory of Chapter 22’s weekend problem enters linear algebra.

Example 20.20

In KnK^n, a hyperplane is a solution set {a1x1+⋯+anxn=0}\{a_1 x_1 + \dots + a_n x_n = 0\} with not all aia_i zero — the familiar equation of a plane through the origin in R3\R^3. In function spaces, evaluation forms P↦P(1)P \mapsto P(1) or f↦∫01ff \mapsto \int_0^1 f define hyperplanes of Rn[X]\R_n[X], of C([0, 1])C(\intcc{0}{1}) (cf. Exercise 19.6).

Example 20.21 (One hyperplane, worked three ways)

Take φ(x,y,z)=x−2y+3z\varphi(x, y, z) = x - 2y + 3z on R3\R^3 and H=ker⁡φH = \ker\varphi. Basis: solve x=2y−3zx = 2y - 3z:

(2y−3z, y, z)=y (2,1,0)+z (−3,0,1),(2y - 3z,\ y,\ z) = y\,(2, 1, 0) + z\,(-3, 0, 1),

two free vectors: dim⁡H=2\dim H = 2, a hyperplane, as Theorem 20.19 predicts from φ≠0\varphi \neq 0. Supplementary line: any vector outside HH spans one, e.g. a=(1,0,0)a = (1, 0, 0) (φ(a)=1≠0\varphi(a) = 1 \neq 0); the decomposition of an arbitrary vv is explicit:

v=(v−φ(v) a)⏟∈ H+φ(v) a⏟∈ Vect⁡(a),v = \underbrace{\bigl(v - \varphi(v)\,a\bigr)}_{\in\,H} + \underbrace{\varphi(v)\,a}_{\in\,\operatorname{Vect}(a)},

since φ(v−φ(v)a)=φ(v)−φ(v)φ(a)=0\varphi\bigl(v - \varphi(v)a\bigr) = \varphi(v) - \varphi(v)\varphi(a) = 0. Proportionality: if ψ(x,y,z)=−2x+4y−6z\psi(x,y,z) = -2x + 4y - 6z, then ψ=−2φ\psi = -2\varphi and both have kernel HH; conversely, any form vanishing on HH is a multiple of φ\varphi (Exercise 20.8) — the equation of a hyperplane is unique up to scale, a fact used constantly for planes in geometry.

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