8.1 Dot Products and Orthogonal Sets

Definition: Suppose that u = (u₁, …, u_n) and v = (v₁, …, v_n) are both vectors in ℝⁿ. Then the dot product of u and v is u ⋅ v = u₁v₁ + … + u_nv_n.

Theorem: Let u, v, w be in ℝⁿ. Then the dot product has the following properties:

• (Symmetry) u ⋅ v = v ⋅ u,
• (Linearity) (cu + v) ⋅ w = cu ⋅ w + v ⋅ w,
• (Positive Definite) u ⋅ u ≥ 0 for all u, and u ⋅ u = 0 if and and only if u = 0.

Definition: Let x be a vector in ℝⁿ, then the norm of x is given by $\|x\|=\sqrt{x\cdot x}$. Note that ∥cx∥ = |c|∥x∥.

For two vectors u and v, the distance between u and v is given by ∥u − v∥.

Definition: Let u and v be vectors in ℝⁿ are orthogonal if u ⋅ v = 0.

Theorem: (Pythagorean Theorem) Suppose that u and v are in ℝⁿ. Then ∥u + v∥² = ∥u∥² + ∥v∥² if and only if u ⋅ v = 0.

Theorem: (Triangle Inequality) If u and v are in ℝⁿ, then ∥u + v∥ ≤ ∥u∥ + ∥v∥.

Definition: Let S be a subspace of ℝⁿ. A vector u is orthogonal to S if it is orthogonal to every vector in S. The set of all vectors orthogonal to S is called the orthogonal complement of S and is denoted S^⊥.

The orthogonal complement to a subspace is also a subspace.

Theorem: Let B = {v₁, …, v_n} be a basis for a subspace S. Then u ∈ S^⊥ (u is orthogonal to S) if and only if u is orthogonal to each v_i.

Example: Let s₁ = (1, 0,  − 1) and s₂ = (1, 1, 1) and S be the span of s₁ and s₂. Is u = ( − 1, 1, 1) ∈ S^⊥? What is a basis for S^⊥?

Definition: A set of vectors V in ℝⁿ form an orthogonal set the vectors are pairwise orthogonal. This means that if v_i and v_j are distinct vectors in V, then v_i ⋅ v_j = 0.

Example:

• Is the standard basis an orthogonal set?
• What’s a basis that is not orthogonal?

Theorem: An orthogonal set of nonzero vectors is linearly independent.

Definition: A basis that is orthogonal as a set is called an orthogonal basis. A basis that is orthogonal as a set and is comprised of vectors of norm 1 is called an orthonormal basis.

Theorem: Let S be a subspace with orthogonal basis {v₁, …, v_k}. Then any vector s ∈ S can be written as a linear combination v = c₁v₁ + … + c_kv_k with c_i = v_i ⋅ s/∥v_i∥².

Example: (THIS IS A BAD EXAMPLE. TURNS OUT NOT TO BE ORTHOGONAL.) Let v₁ = ( − 2, 1, 1), v₂ = (1,  − 1,  − 3), v₃ = (4, 7,  − 1). Write (3,  − 1, 5) as a linear combination of v_i.

For finite dimensional spaces, we have that (S^⊥)^⊥ = S. Use this to show that every subspace is the null space of some matrix.