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Lesson 1: Dot Product, Length, and Orthogonality

Estimated time: 40-50 minutes

Learning Objectives

The Dot Product

Dot Product: For u = (u1, u2, ..., un) and v = (v1, v2, ..., vn) in R^n:

u · v = u1*v1 + u2*v2 + ... + un*vn = u^T v.

Worked Example 1

u = (2, -1, 3), v = (4, 0, -1). u · v = 2(4) + (-1)(0) + 3(-1) = 8 + 0 - 3 = 5.

Properties of the Dot Product: For vectors u, v, w and scalar c:

  • u · v = v · u (commutative)
  • (u + v) · w = u · w + v · w (distributive)
  • (cu) · v = c(u · v)
  • u · u ≥ 0, and u · u = 0 iff u = 0 (positive definite)

Vector Length and Distance

Length (Norm): ||u|| = sqrt(u · u) = sqrt(u1^2 + u2^2 + ... + un^2).

Unit Vector: A vector with ||u|| = 1. To normalize: u-hat = u / ||u||.

Distance: dist(u, v) = ||u - v||.

Worked Example 2

u = (3, -4). ||u|| = sqrt(9 + 16) = sqrt(25) = 5.

Unit vector: u-hat = (3/5, -4/5). Check: ||(3/5, -4/5)|| = sqrt(9/25 + 16/25) = 1.

Worked Example 3

u = (1, 2, 3), v = (4, 0, 1). dist(u,v) = ||(1-4, 2-0, 3-1)|| = ||(-3, 2, 2)|| = sqrt(9+4+4) = sqrt(17).

Orthogonality

Orthogonal Vectors: u and v are orthogonal (perpendicular) if u · v = 0. We write u ⊥ v.

Worked Example 4

u = (2, 1, -1), v = (1, -3, -1). u · v = 2 - 3 + 1 = 0. So u and v are orthogonal.

Pythagorean Theorem: If u ⊥ v, then ||u + v||^2 = ||u||^2 + ||v||^2.

The Angle Between Vectors

cos(theta) = (u · v) / (||u|| * ||v||). Vectors are orthogonal when theta = 90 degrees (cos theta = 0).

The Cauchy-Schwarz Inequality

Cauchy-Schwarz: |u · v| ≤ ||u|| * ||v|| for all u, v. Equality holds iff u and v are parallel (one is a scalar multiple of the other).

Worked Example 5

u = (1, 2), v = (3, 1). |u · v| = |3 + 2| = 5. ||u||*||v|| = sqrt(5)*sqrt(10) = sqrt(50) = 5*sqrt(2) approximately 7.07.

Check: 5 ≤ 7.07. Cauchy-Schwarz holds, and they are not parallel.

Orthogonal and Orthonormal Sets

Orthogonal Set: A set of vectors {v1, v2, ..., vk} where vi · vj = 0 for all i not equal to j.

Orthonormal Set: An orthogonal set where every vector has unit length: ||vi|| = 1.

Worked Example 6

S = {(1, 0, 0), (0, 1, 0), (0, 0, 1)} -- the standard basis. Every pair has dot product 0, and each has length 1. This is orthonormal.

S = {(1, 1, 0), (-1, 1, 0), (0, 0, 3)}. Check: (1,1,0)·(-1,1,0) = -1+1 = 0. (1,1,0)·(0,0,3) = 0. (-1,1,0)·(0,0,3) = 0. Orthogonal but not orthonormal (lengths are sqrt(2), sqrt(2), and 3).

Key Property: An orthogonal set of nonzero vectors is automatically linearly independent. This makes orthogonal bases especially useful.

Coordinates in an Orthogonal Basis

If {v1, ..., vn} is an orthogonal basis, then any vector x can be written as:

x = [(x · v1)/(v1 · v1)] v1 + [(x · v2)/(v2 · v2)] v2 + ... + [(x · vn)/(vn · vn)] vn

No row reduction needed! For an orthonormal basis, this simplifies to x = (x · v1)v1 + (x · v2)v2 + ... + (x · vn)vn.

Worked Example 7

Orthogonal basis: v1 = (1, 1), v2 = (1, -1). Express x = (5, 3).

c1 = (x · v1)/(v1 · v1) = (5+3)/2 = 4. c2 = (x · v2)/(v2 · v2) = (5-3)/2 = 1.

x = 4(1,1) + 1(1,-1) = (4,4) + (1,-1) = (5, 3). Correct.

Check Your Understanding

1. Compute (2, -3, 1) · (1, 4, -2).

Answer: 2(1) + (-3)(4) + 1(-2) = 2 - 12 - 2 = -12.

2. Find the length of v = (1, -2, 2, 0).

Answer: ||v|| = sqrt(1 + 4 + 4 + 0) = sqrt(9) = 3.

3. Are (3, -1) and (1, 3) orthogonal?

Answer: (3)(1) + (-1)(3) = 3 - 3 = 0. Yes, they are orthogonal.

4. Find the angle between u = (1, 0) and v = (1, 1).

Answer: cos(theta) = (1+0)/(1*sqrt(2)) = 1/sqrt(2). theta = 45 degrees (pi/4 radians).

Key Takeaways

Next Lesson

Orthogonal Projections.

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