← Module 02 / Vectors & spaces

Lesson 02.1 · §3.1

What can you build from two directions?

Explain how span, independence, and a basis fit together.

In this lesson 3 ideas
  1. Build a target
  2. Reach and redundancy
  3. Transfer

Source scope: §3.1. Worked numerical examples, experiments, and practice are teaching supplements unless identified as source results.

Build a target

Suppose two adjustable ingredients contribute u=(1,0)u=(1,0) and v=(1,2)v=(1,2). To build (3,2)(3,2), take two copies of uu and one of vv: 2u+v=(3,2)2u+v=(3,2). The coefficients may be negative or zero. We are describing combinations, not quantities you must buy.

A real vector space is a set VV with addition V×V→VV\times V\to V and scalar multiplication R×V→V\mathbb R\times V\to V. These operations stay in VV. For every x,y,z∈Vx,y,z\in V and α,β∈R\alpha,\beta\in\mathbb R, the source lists these axioms:

  1. An additive identity 0∈V0\in V satisfies x+0=xx+0=x.
  2. Each xx has an additive inverse −x-x, with x+(−x)=0x+(-x)=0.
  3. The scalar identity satisfies 1x=x1x=x.
  4. Addition commutes: x+y=y+xx+y=y+x.
  5. Addition and scaling associate: (x+y)+z=x+(y+z)(x+y)+z=x+(y+z) and α(βx)=(αβ)x\alpha(\beta x)=(\alpha\beta)x.
  6. Both distributive laws hold: α(x+y)=αx+αy\alpha(x+y)=\alpha x+\alpha y and (α+β)x=αx+βx(\alpha+\beta)x=\alpha x+\beta x.

Source footnote 1: Vector spaces can use any field of scalars. These notes use the real field to avoid a detour into abstract algebra.

Reach and redundancy

A linear combination is ∑i=1kαivi\sum_{i=1}^k\alpha_i v_i. The span is the set of all such combinations:

span⁡{v1,…,vk}={v∈V:∃α1,…,αk∈R, v=∑iαivi}.\operatorname{span}\{v_1,\ldots,v_k\}=\{v\in V:\exists\alpha_1,\ldots,\alpha_k\in\mathbb R,\ v=\textstyle\sum_i\alpha_i v_i\}.

The vectors are linearly independent when

∑iαivi=0⟹α1=⋯=αk=0.\sum_i\alpha_i v_i=0\quad\Longrightarrow\quad\alpha_1=\cdots=\alpha_k=0.

Our two ingredients are independent: the second coordinate of au+bv=0au+bv=0 gives 2b=02b=0, then the first gives a=0a=0. They also span the plane: (s,t)=(s−t/2)u+(t/2)v(s,t)=(s-t/2)u+(t/2)v. A basis is both independent and spanning. Every independent list is a basis for its own span.

If a finite list spans VV, the space is finite-dimensional. Its dimension dim⁡V\dim V is the number of vectors in a basis. Otherwise it is infinite-dimensional. As a teaching example, all real polynomials form an infinite-dimensional space: no finite list can generate polynomials of arbitrarily high degree.

Explore it yourself

Build a vector from two ingredients

Predict whether changing the second direction can make every point reachable.

Preparing interactive controls…

x₁x₂-4-4-3-3-2-2-1-1011223344uva ub vsum

u = (1.000, 0); v = (1.000, 2.000). a u + b v = (2.000, 2.000). The directions are independent. Unrestricted coefficients reach any point in the plane; the sliders show a bounded sample.

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

Transfer

Can (1,2)(1,2) and (2,4)(2,4) form a basis for the plane?

Explain the obstruction

No. Twice the first vector minus the second is zero, so the list is dependent. Every combination has second coordinate twice its first. It spans a line, not the plane. The single nonzero vector (1,2)(1,2) is a basis for that line.