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Vector Spaces

Source lecture(s): PHY621 Ch. 1

Intuition

The course opens with an act of abstraction that pays for itself for the rest of your career: stop thinking of a vector as an arrow. A vector is any object you can add to another of its kind and scale by a number, with the results behaving sensibly. Arrows qualify. So do polynomials, matrices, solutions of a linear ODE, and quantum states.

Once you notice that, every theorem proved about arrows applies unchanged to all of them. This is why one chapter of linear algebra underpins the whole of physics.

The axioms

A set \(V\) with addition and scalar multiplication is a vector space if, for all \(\mathbf{u},\mathbf{v}\in V\) and scalars \(\alpha,\beta\):
closure \(\mathbf{u}+\mathbf{v}\in V\), \(\alpha\mathbf{u}\in V\)
commutative, associative \(\mathbf{u}+\mathbf{v} = \mathbf{v}+\mathbf{u}\), \((\mathbf{u}+\mathbf{v})+\mathbf{w} = \mathbf{u}+(\mathbf{v}+\mathbf{w})\)
identity, inverse \(\exists\,\mathbf{0}\) with \(\mathbf{u}+\mathbf{0} = \mathbf{u}\); \(\exists\,-\mathbf{u}\)
distributive \(\alpha(\mathbf{u}+\mathbf{v}) = \alpha\mathbf{u}+\alpha\mathbf{v}\), \((\alpha+\beta)\mathbf{u} = \alpha\mathbf{u}+\beta\mathbf{u}\)
scalar identity \(1\cdot\mathbf{u} = \mathbf{u}\)

Notice what is not here: no length, no angle, no dot product. Those come later with the inner product, and they are extra structure — a vector space by itself has no notion of "how long" a vector is.

Span, independence, basis

  • Span: all linear combinations of a set. The span of \(\{\mathbf{v}_1,\dots\}\) is the smallest subspace containing them.
  • Linear independence: \(\sum c_i\mathbf{v}_i = 0\) only when every \(c_i = 0\). Equivalently, no vector in the set is a combination of the others — nothing is redundant.
  • Basis: an independent spanning set. Every vector then has a unique expansion \(\mathbf{v} = \sum v_i\mathbf{e}_i\), and the \(v_i\) are its components in that basis.

The number of basis vectors is the dimension, and it does not depend on which basis you pick.

Why physicists care about the last point

Components are not the vector. A vector is a geometric object; its components are a description relative to a chosen basis, and they change when you rotate the basis while the vector does not. Almost every "is this quantity physical?" question in the course reduces to: does it transform correctly under a change of basis? See orthogonal transformations.

Examples worth carrying around:

Vector space Vectors are Dimension
\(\mathbb{R}^3\) arrows 3
Polynomials of degree \(\le n\) \(1, x, x^2, \dots\) \(n+1\)
Solutions of \(y'' + \omega^2 y = 0\) \(\sin\omega x\), \(\cos\omega x\) 2
Square-integrable functions \(L^2\) wavefunctions infinite

The third is why "the general solution is a linear combination of two independent solutions" is a theorem rather than a lucky observation — the solution set of a linear homogeneous ODE is a vector space, of dimension equal to the order. The fourth is quantum mechanics.

Common mistakes

  • Assuming a vector space has lengths and angles. It does not until you supply an inner product. Different inner products on the same space give different geometries — and in the normal modes lab the physically correct one is mass-weighted.
  • Confusing a vector with its components. The components change under a change of basis; the vector does not.
  • Thinking dimension depends on the basis. It does not; that is a theorem.

Knowledge graph position

Prerequisites: none — this is the foundation. Leads to: inner products, matrices, eigenproblems, and every transform in the course.

Quiz

Q1 (conceptual). Why is the solution set of \(y'' + \omega^2 y = 0\) a vector space, and what is its dimension?

Answer

The equation is linear and homogeneous, so any linear combination of solutions is a solution — which is exactly closure under addition and scaling. Its dimension is 2 (the order of the ODE), with basis \(\{\sin\omega x, \cos\omega x\}\). "The general solution is \(A\sin + B\cos\)" is then a statement about a basis, not a guess.

Q2 (conceptual). Do the vectors \(\{(1,2),(2,4),(0,1)\}\) form a basis of \(\mathbb{R}^2\)?

Answer

No. They span \(\mathbb{R}^2\), but they are not independent — \((2,4) = 2(1,2)\) — and there are three of them in a 2-dimensional space, so they cannot be. Any two of them that are not parallel do form a basis.

Q3 (MCQ). Which is not guaranteed by the vector-space axioms alone?

  • (a) that every vector has an additive inverse
  • (b) that a unique expansion exists in a given basis
  • (c) that the length of a vector is defined
  • (d) that scalar multiplication distributes over addition
Answer

(c). Length requires a norm, which usually comes from an inner product — extra structure beyond the axioms. (a), (b) and (d) all follow from them.