Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16
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Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors

Statement

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

∣⟨u,v⟩∣≤∥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).

Proof

technique · direct
1.1L1L2L3

If v=0, both sides are zero and the pair is dependent. Suppose v≠0, put c=⟨u,v⟩/⟨v,v⟩, and use [L1] to expand 0≤⟨u−cv,u−cv⟩=∥u∥2−∣⟨u,v⟩∣2/∥v∥2.

1.2L1L2L3L4

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

2.1step 1.1L2L3algebra

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

3.1step 1.1step 2.1L1L4

Under v≠0, equality in step 2.1 holds exactly when ⟨u−cv,u−cv⟩=0, which by [L1] is exactly u=cv. Thus equality implies dependence. The already separated case v=0 does too.

4.1step 1.2step 2.1step 3.1∎

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

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources