October 4

Announcements

• Section 1.1, 1.2 due tomorrow
• Worksheet 1 due Friday - there are html issues, see pdf version
• Section 2.1, 2.2 due next Thursday
• Office hours today after class and tomorrow 12-1

Linear combinations and span

Definition: If u₁, u₂, …, u_m are vectors and c₁, c₂, …, c_m are scalars, then
c₁u₁ + c₂u₂ + … + c_mu_m
is a linear combination of u₁, …, u_m. Note that the constants can be negative or zero.

Definition: Let S = {u₁, u₂, …, u_m} be a set of vectors. Then the span of S, spanS, is the set of all linear combinations of u₁, u₂, …, u_m}.

What vectors in ℝ² are a linear combination of (1, 0) and (0, 1)? In other words, what vectors are in the span of (1, 0) and (0, 1)?

What vectors in ℝ² are a linear combination of (1, 2) and (0, 1)? Talk about lines and averages here.

Is (3, 4) a linear combination of (1, 2) and (0, 1)? In other words, is (3, 4) in the span of (1, 2) and (0, 1)? In other words, does there exists x₁, x₂ ∈ ℝ such that x₁(1, 2) + x₂(0, 1) = (3, 4)? In other words, system of equations!

Every system of equation can be interpeted in this way.

Theorem: Let u₁, …, u_m and v be vectors in ℝⁿ. Then v ∈ span({u₁, …, u_m}) if and only if the linear system with augmented matrix [u₁ u₂ … u_m|v] has a solution.

The solution space can be expressed as a linear combination.

Theorem: Let u₁, u₂, …, u_m be vectors in ℝⁿ. If u∈span({u₁, …, u_m}), then span({u₁, …, u_m})=span({u₁, …, u_m, u}).

When does a set of vectors span ℝⁿ?

Theorem: Let u₁, u₂, …, u_m be vectors in ℝⁿ. Let A = [u₁ u₂ … u_m] and B ∼ A, where B is in echelon form. Then span({u₁, …, u_m}) = ℝⁿ if and only if B has a pivot position in every row.

give outline of proof

We can write linear systems as Ax = b.