Inner Product: How Much Overlap

How much do two vectors point the same way? The inner product is the rule that measures the overlap.

It keeps three promises. Swapping the two sides changes nothing. Sums and scalings pull straight through. And the inner product of a vector with itself is the square of its length. So an inner product brings a norm with it. An inner product of 0 means the vectors are perpendicular.

For example, two people push a 📦 box. Pushing in the same direction, their forces add up in full. Pushing at a right angle, the sideways force does nothing for forward motion. The inner product measures exactly the part that adds up.

⟨x,y⟩=x1y1+x2y2\langle x, y \rangle = x_1 y_1 + x_2 y_2
⟨x,x⟩=∥x∥2\langle x, x \rangle = \|x\|^2
⟨x,y⟩=∥x∥ ∥y∥cos⁡θ\langle x, y \rangle = \|x\| \, \|y\| \cos\theta

Examples & Exercises

For x=(1,2)x = (1, 2) and y=(3,−1)y = (3, -1), compute ⟨x,y⟩\langle x, y \rangle, ∥x∥\|x\|, ∥y∥\|y\|, and the cosine of the angle between them. Are the vectors perpendicular?

⟨x,y⟩=x1y1+x2y2,cos⁡θ=⟨x,y⟩∥x∥ ∥y∥\langle x, y \rangle = x_1 y_1 + x_2 y_2, \qquad \cos\theta = \frac{\langle x, y \rangle}{\|x\| \, \|y\|}
Exactly the two vectors from the problem.
Show proof

  1. ⟨x,y⟩=1⋅3+2⋅(−1)=3−2=1\langle x, y \rangle = 1 \cdot 3 + 2 \cdot (-1) = 3 - 2 = 1.

  2. ∥x∥=12+22=5\|x\| = \sqrt{1^2 + 2^2} = \sqrt{5} and ∥y∥=32+(−1)2=10\|y\| = \sqrt{3^2 + (-1)^2} = \sqrt{10}.

  3. cos⁡θ=15⋅10=150=152≈0.1414\cos\theta = \frac{1}{\sqrt{5} \cdot \sqrt{10}} = \frac{1}{\sqrt{50}} = \frac{1}{5\sqrt{2}} \approx 0.1414. The inner product is not 0, so the vectors are not perpendicular. It is positive, so the angle is acute.

Prove completely that the standard inner product ⟨x,y⟩=x1y1+x2y2\langle x, y \rangle = x_1 y_1 + x_2 y_2 on R2\mathbb{R}^2 satisfies the three inner-product axioms: symmetry, linearity in one slot, and positive definiteness.

⟨x,y⟩=x1y1+x2y2\langle x, y \rangle = x_1 y_1 + x_2 y_2
e1,e2e_1, e_2 — a perpendicular pair with inner product 0.
Show proof

  1. Symmetry: multiplication of reals commutes, so ⟨y,x⟩=y1x1+y2x2=x1y1+x2y2=⟨x,y⟩\langle y, x \rangle = y_1 x_1 + y_2 x_2 = x_1 y_1 + x_2 y_2 = \langle x, y \rangle.

  2. Linearity: for x,x′,yx, x', y and real aa, ⟨ax+x′,y⟩=(ax1+x1′)y1+(ax2+x2′)y2=a(x1y1+x2y2)+(x1′y1+x2′y2)=a⟨x,y⟩+⟨x′,y⟩\langle a x + x', y \rangle = (a x_1 + x_1') y_1 + (a x_2 + x_2') y_2 = a (x_1 y_1 + x_2 y_2) + (x_1' y_1 + x_2' y_2) = a \langle x, y \rangle + \langle x', y \rangle. Just the distributive law.

  3. Positive definiteness: ⟨x,x⟩=x12+x22≥0\langle x, x \rangle = x_1^2 + x_2^2 \ge 0, and the sum of squares is 0 only when x1=x2=0x_1 = x_2 = 0. All three axioms hold, so the standard inner product is an inner product.

Prove completely that in a real inner product space, the polarization identity ⟨x,y⟩=∥x+y∥2−∥x−y∥24\langle x, y \rangle = \frac{\|x + y\|^2 - \|x - y\|^2}{4} holds.

⟨x,y⟩=∥x+y∥2−∥x−y∥24\langle x, y \rangle = \frac{\|x + y\|^2 - \|x - y\|^2}{4}
x=(1,1)x = (1, 1), y=(1,0)y = (1, 0) — ∥x+y∥2=8\|x + y\|^2 = 8 and ∥x−y∥2=4\|x - y\|^2 = 4, so the inner product is 1.
Show proof

  1. Expand the norms through the inner product: ∥x+y∥2=⟨x+y,x+y⟩=∥x∥2+2⟨x,y⟩+∥y∥2\|x + y\|^2 = \langle x + y, x + y \rangle = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2, and likewise ∥x−y∥2=∥x∥2−2⟨x,y⟩+∥y∥2\|x - y\|^2 = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2.

  2. Subtracting, the norm terms cancel: ∥x+y∥2−∥x−y∥2=4⟨x,y⟩\|x + y\|^2 - \|x - y\|^2 = 4\langle x, y \rangle.

  3. Dividing by 4 gives the polarization identity. The inner product is fully recovered from the norm alone — the norm determines the inner product.

3. Sequences