← Module 02 / Vectors & spaces

Lesson 02.10 · §3.5.1–3.5.2

How large can an inner product be?

Prove the orthogonal length identity and identify Cauchy–Schwarz equality cases.

In this lesson 3 ideas
  1. When do squared lengths add?
  2. Alignment has a limit
  3. Transfer

Source scope: §3.5.1–3.5.2. Worked numerical examples, experiments, and practice are teaching supplements unless identified as source results.

When do squared lengths add?

For (3,0)(3,0) and (0,4)(0,4), the sum has squared length 25=9+1625=9+16. For (3,0)(3,0) and (4,0)(4,0), it has squared length 4949. The missing condition is orthogonality.

Theorem 1 (source). In a real inner product space, if x⊥yx\perp y, then

∥x+y∥2=∥x∥2+∥y∥2.\|x+y\|^2=\|x\|^2+\|y\|^2.

Source proof. Expand using linearity and symmetry:

∥x+y∥2=⟨x+y,x+y⟩=⟨x,x⟩+⟨y,x⟩+⟨x,y⟩+⟨y,y⟩=∥x∥2+∥y∥2.\|x+y\|^2=\langle x+y,x+y\rangle=\langle x,x\rangle+\langle y,x\rangle+\langle x,y\rangle+\langle y,y\rangle=\|x\|^2+\|y\|^2.

Both cross terms vanish because the vectors are orthogonal. This proof works in any real inner product space, not just a drawn triangle.

Alignment has a limit

The Cauchy–Schwarz inequality says

∣⟨x,y⟩∣≤∥x∥∥y∥.|\langle x,y\rangle|\le\|x\|\|y\|.

Equality holds exactly when the pair is linearly dependent. This includes the case where either vector is zero. For two nonzero vectors it means that one is a scalar multiple of the other. This wording also covers a zero vector paired with a nonzero vector without requiring the nonzero vector to be a multiple of zero.

The source states the inequality and equality condition without proof. Here is a teaching proof. If y=0y=0, both sides are zero. Otherwise set a=⟨x,y⟩/∥y∥2a=\langle x,y\rangle/\|y\|^2. Positivity gives

0≤∥x−ay∥2=∥x∥2−⟨x,y⟩2∥y∥2.0\le\|x-ay\|^2=\|x\|^2-\frac{\langle x,y\rangle^2}{\|y\|^2}.

Multiply by the positive denominator and take square roots. Equality holds precisely when x−ay=0x-ay=0. For (1,2)(1,2) and (2,4)(2,4), both sides equal 1010. For (1,0)(1,0) and (0,1)(0,1), the left side is zero and the right side is one.

Explore it yourself

Find the closest point on a line

Move the candidate along S. Can its distance beat the perpendicular foot?

Preparing interactive controls…

x₁x₂-4-4-3-3-2-2-1-1011223344SvPvresiduals

v = (2, 1); Pv = (1.933, 1.116); residual = (0.067, -0.116). Residual · u = 0. Candidate distance² = 1.536 = residual length² 0.018 + foot-to-candidate length² 1.518.

Axes: first and second real coordinates, in unit steps. Endpoints and labels identify each vector.

Transfer

Can the absolute dot product of vectors with lengths two and three equal seven?

Apply the bound before calculating coordinates

No. The bound is six. Equality at six requires a shared line, with either the same or opposite direction.