core18 min studyLesson 11 in Mathematics

Vector spaces and bases

Separate a vector from its coordinates, and discover what a basis actually guarantees.

On this page

What you’ll understand

  • Test whether a proposed set is a vector subspace.
  • Distinguish spanning, independence, and a basis.
  • Recover coordinates in an unfamiliar two-dimensional basis.

A pair of numbers can describe a displacement, two sensor readings, or coefficients of a polynomial. The useful common structure is the ability to add objects and scale them consistently. A vector space captures that structure without insisting that every vector be a physical arrow. This distinction becomes essential when quantum states are functions and machine-learning inputs have thousands of features: the coordinate list is a description of an object relative to a chosen basis.

A space must survive the allowed operations

Work over the real numbers first. A vector space has an addition operation and multiplication by real scalars, obeying the familiar rules for associativity, distributivity, a zero vector, and additive inverses. A subspace of an existing vector space contains zero and is closed under addition and scalar multiplication. The line y=2xy=2x passes these tests. The parallel line y=2x+1y=2x+1 fails immediately because it excludes the origin.

The objects need not be arrows. Polynomials of degree at most two form a three-dimensional real vector space: adding two such polynomials or scaling one never introduces a cubic term. Polynomials of degree exactly two do not form a space, since subtracting a polynomial from itself produces zero. Asking what happens under cancellation is often faster than trying to memorize a catalogue of examples.

span⁡(v1,…,vk)={∑i=1kaivi:ai∈R}\operatorname{span}(v_1,\ldots,v_k)=\left\{\sum_{i=1}^{k}a_i v_i:a_i\in\mathbb R\right\}
For vectors in a real vector space, the span contains every finite linear combination, including zero when all coefficients vanish.↔ Scroll wide equations horizontally.

A basis gives existence and uniqueness

A collection spans a space if it can express every vector in that space. It is linearly independent if the only combination producing zero has every coefficient equal to zero. A basis satisfies both conditions. Spanning guarantees that coordinates exist; independence guarantees that they are unique. Three arrows in a plane can span the plane, but their redundant descriptions cannot provide a unique three-coordinate address.

The number of vectors in a basis is the dimension of a finite-dimensional space. The standard basis of the plane is convenient, not mandatory. Choosing two tilted, nonparallel arrows changes the coordinate numbers without changing the underlying displacement. When checking independence, keep the ambient space explicit: two independent vectors form a basis of their span, but they cannot span all of three-dimensional space.

v=∑i=1ncibi,∑i=1naibi=0⟹a1=⋯=an=0v=\sum_{i=1}^{n}c_i b_i,\qquad \sum_{i=1}^{n}a_i b_i=0\Longrightarrow a_1=\cdots=a_n=0
If the finite collection b₁,…,bₙ spans the space and obeys the displayed independence condition, every vector has exactly one coordinate tuple c.↔ Scroll wide equations horizontally.

Geometry is extra structure

A vector space alone does not specify angles or lengths. An inner product supplies that additional structure. For ordinary real coordinate vectors, the dot product produces lengths and tests orthogonality. An orthonormal basis consists of mutually perpendicular unit vectors; it makes each coordinate a simple dot product. A general basis is still valid, but coordinates then require solving simultaneous equations rather than merely reading projections.

This distinction prevents two later confusions. Quantum amplitudes use complex inner-product spaces, so conjugation enters the geometry. Machine-learning features can be expressed in any basis, but distances and regularization depend on the metric and scaling selected. We are establishing finite-dimensional language here; completeness, infinite-dimensional function spaces, and the quantum measurement rule require additional lessons rather than following automatically from an arrow drawing.

Give one displacement a different address

Express v=(4,2)v=(4,2) in the basis b1=(1,1)b_1=(1,1) and b2=(1,−1)b_2=(1,-1).

  1. Write the representation before guessing coefficients: v=c1b1+c2b2v=c_1b_1+c_2b_2, so its ordinary coordinates are (c1+c2,c1−c2)(c_1+c_2,c_1-c_2).

  2. Match components to obtain c1+c2=4c_1+c_2=4 and c1−c2=2c_1-c_2=2. Adding eliminates c2c_2, giving 2c1=62c_1=6 and hence c1=3c_1=3.

  3. Substitute into the first equation to get c2=1c_2=1. Reconstruct 3(1,1)+(1,−1)=(4,2)3(1,1)+(1,-1)=(4,2) to check both components.

The result

The new coordinates are (3,1). The displacement remains (4,2) in the standard basis; no physical motion occurred when its coordinate description changed.

Explore the model

Interactive lab

Interactive lab · vectors

Resolve a vector in a fixed basis

Change angle and magnitude to inspect one vector and its x/y projections in the standard basis. Pause and predict the component signs before stepping. The projections supply coordinates in this orthonormal basis; the lab does not change to arbitrary tilted bases or test independence.

Resolve a vector in a fixed basisPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forOne vector has coordinates relative to a chosen basis; its components depend on both direction and magnitude.

Check your understanding

Why is the set of points satisfying x+y=1 not a vector subspace of the plane?

Try it yourself

Are (1,2)(1,2) and (2,4)(2,4) a basis for R2\mathbb R^2? Describe their span.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Ask whether one vector is a scalar multiple of the other, then inspect the zero combination.

Compare with the solution

They are dependent because 2(1,2)−(2,4)=02(1,2)-(2,4)=0 uses nonzero coefficients. Their span is only the line y=2xy=2x, so they neither supply unique two-coordinate descriptions nor reach the whole plane. A single nonzero vector from the pair is a basis of that line.

Take it with you

  • A subspace is closed under linear combinations and includes zero.
  • A basis spans without redundancy, giving unique coordinates.
  • Lengths and angles require an inner product in addition to vector-space operations.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.