TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors

Statement

For vectors $u, v$ in a real or complex inner product space,

$∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥ .$

Equality holds if and only if $u$ and $v$ are linearly dependent, including the case in which either vector is zero.

Facts & Assumptions

Given: Vectors $u, v$ in an inner product space over $R$ or $C$ .

[L1]

The inner product is linear in the first variable, conjugate-linear in the second, conjugate symmetric, and positive definite (Real and complex inner product spaces, with the inner product linear in the first argument).

[L2]

The norm is the nonnegative square root of the diagonal pairing (The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ).

[L3]

Complex modulus satisfies $z \overline{z} = ∣ z ∣^{2}$ and vanishes exactly at zero (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L4]

A two-vector list is dependent exactly when a nontrivial scalar combination vanishes (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

Proof

technique · direct

1.1

If $v = 0$ , both sides are zero and the pair is dependent. Suppose $v \neq 0$ , put $c = ⟨ u, v ⟩ / ⟨ v, v ⟩$ , and use [L1] to expand $0 \leq ⟨ u - c v, u - c v ⟩ = ∥ u ∥^{2} - ∣ ⟨ u, v ⟩ ∣^{2} / ∥ v ∥^{2}$ .

L1L2L3

1.2

Conversely, if $u, v$ are dependent and neither is zero, write $u = c v$ ; then $∣ ⟨ u, v ⟩ ∣ = ∣ c ∣ ∥ v ∥^{2} = ∥ u ∥ ∥ v ∥$ . If either is zero, equality is immediate.

L1L2L3L4

2.1

Multiplying step 1.1 by the positive number $∥ v ∥^{2}$ gives $∣ ⟨ u, v ⟩ ∣^{2} \leq ∥ u ∥^{2} ∥ v ∥^{2}$ . Since both sides of the desired inequality are nonnegative, factoring the difference of their squares gives the stated inequality.

step 1.1L2L3algebra

3.1

Under $v \neq 0$ , equality in step 2.1 holds exactly when $⟨ u - c v, u - c v ⟩ = 0$ , which by [L1] is exactly $u = c v$ . Thus equality implies dependence. The already separated case $v = 0$ does too.

step 1.1step 2.1L1L4

4.1

Steps 2.1, 3.1, and 1.2 prove the inequality and both equality directions.

step 1.2step 2.1step 3.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 86 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., §6A (standard reference, not scraped)
Sergei Treil, Linear Algebra Done Wrong, Ch. 5, §5.1 (standard reference, not scraped)

Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors