CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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The induced length is a norm

Statement

Let $V$ be a real or complex inner-product space with induced length $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ . Then $∥ v ∥ \geq 0$ with $∥ v ∥ = 0$ exactly for $v = 0$ , $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ for every scalar $λ$ , and $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ ; hence $∥ \cdot ∥$ is a norm on $V$ , read over $R$ by A norm on a real vector space, the induced metric, and the dictionary with the metric axioms and over $C$ by Complex norm and complex normed space.

Facts & Assumptions

[A1]

The induced length is the unique nonnegative square root of the diagonal pairing, with $∥ v ∥ \geq 0$ and $∥ v ∥ = 0$ exactly for $v = 0$ (The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product).

[A2]

Cauchy–Schwarz gives $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ (Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs).

[A3]

A norm on a real vector space satisfies separation, absolute homogeneity and the triangle inequality, and a complex normed space is defined by the same clauses with the complex modulus (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms, Complex norm and complex normed space).

[A4]

For complex scalars $z \overline{z} = ∣ z ∣^{2}$ and $λ \overline{λ} = ∣ λ ∣^{2}$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive), and for real scalars $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ with $∣ x ∣ \geq 0$ (Basic properties of the absolute value).

[A5]

For nonnegative reals $a \leq b$ if and only if $a^{2} \leq b^{2}$ (Squaring is monotone on the nonnegatives), and each nonnegative real has a unique nonnegative square root (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[A6]

If $z = a + b i$ then $Re z = a$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ , so $Re z \leq ∣ z ∣$ (Real and imaginary parts, complex conjugation, and modulus).

Proof

technique · direct

Given: A real or complex inner-product space $V$ , vectors $u, v \in V$ and a scalar $λ$ ; in the real case conjugation is the identity and $∣ \cdot ∣$ is the absolute value of $R$ , so the second clause of [A4] is read in the real field.

1.1A1A3

Nonnegativity and separation are [A1]: $∥ v ∥ \geq 0$ , and $∥ v ∥ = 0$ exactly when $⟨ v, v ⟩ = 0$ , that is exactly when $v = 0$ .

1.2A1A4A5algebra

Homogeneity: $⟨ λ v, λ v ⟩ = λ \overline{λ} ⟨ v, v ⟩ = ∣ λ ∣^{2} ∥ v ∥^{2}$ by sesquilinearity and [A4], and both $∥ λ v ∥$ and $∣ λ ∣ ∥ v ∥$ are nonnegative with equal squares, so $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ by uniqueness of the nonnegative square root.

1.3A1A2A4A6algebra

Triangle inequality: expanding and using conjugate symmetry gives $∥ u + v ∥^{2} = ∥ u ∥^{2} + 2 Re ⟨ u, v ⟩ + ∥ v ∥^{2}$ , and $Re z \leq ∣ z ∣$ for every scalar $z$ , since either $Re z < 0 \leq ∣ z ∣$ or $Re z = a \geq 0$ with $a^{2} \leq a^{2} + b^{2} = ∣ z ∣^{2}$ by [A6]; with [A2] this gives $∥ u + v ∥^{2} \leq ∥ u ∥^{2} + 2 ∥ u ∥ ∥ v ∥ + ∥ v ∥^{2} = (∥ u ∥ + ∥ v ∥)^{2}$ .

2.1A5step 1.3

Both sides of the inequality in step 1.3 are nonnegative, so monotonicity of squaring on the nonnegatives turns it into $∥ u + v ∥ \leq ∥ u ∥ + ∥ v ∥$ .

3.1A3step 1.1step 1.2step 2.1∎

Steps 1.1, 1.2 and 2.1 are exactly the clauses of [A3] over $R$ , and over $C$ they are the same clauses with the complex modulus, so $∥ \cdot ∥$ is a norm on $V$ in either scalar field.

The induced length is a norm

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources