Lecture 1

Sep 1, 2026

Linear transformations, bases, dimension, and coordinate maps

Bases, dimension, and coordinate maps

We begin with a few quick remarks.

Basis: {\begin{matrix} Properties: if it is a basis, then \dots \\ Criteria: if \dots, then it is a basis \\ Calculations: methods \end{matrix}

Dimension of V : {\begin{matrix} Is it unambiguously defined? \end{matrix}

The coordinate map is defined as follows. If $ℬ = (v_{1}, \dots, v_{n})$ is a basis for $V$ , then

\begin{matrix} {[\cdot]}_{ℬ} : V & ⟶ 𝕂^{n}, \\ v & ⟼ [\begin{matrix} c_{1} \\ ⋮ \\ c_{n} \end{matrix}] . \end{matrix}

Definition (Basis and dimension)

A basis $ℬ = (b_{1}, \dots, b_{n})$ of $V$ means that every vector $v \in V$ can be written uniquely as

v = c_{1} b_{1} + \dots + c_{n} b_{n} .

The dimension of $V$ is defined by

\dim V = | ℬ | = n .

We now want to answer questions about the properties and criteria of a basis, and whether dimension is unambiguously defined. These are addressed by the following theorem.

Theorem (Theorem 1 — Theorems 1.4.10 and 1.4.13)

If $ℬ = (b_{1}, \dots, b_{n})$ is a basis of $V$ , then $ℬ$ is linearly independent.
Every finite-dimensional vector space $V$ , meaning $\dim V < \infty$ , has a basis. If $A = (a_{1}, \dots, a_{n})$ is linearly independent and contains $\dim V$ vectors, then $A$ is a basis of $V$ .
The dimension of $V$ is well-defined:
$\dim V = | ℬ | .$
In other words, $\dim V$ is independent of the choice of basis $ℬ$ .

Note (Reminder)

Let $A \in ℝ^{n \times m}$ and

x = {[\begin{matrix} x_{1} & \dots & x_{m} \end{matrix}]}^{T} .

If $m$ , the number of variables, is greater than $n$ , then $A x = 0$ has infinitely many solutions.

Lemma (Lemma 1)

Let $ℬ = (b_{1}, \dots, b_{n})$ be a basis of $V$ , and let $(v_{1}, \dots, v_{m})$ be $m$ vectors in $V$ . If $m > n$ , then $(v_{1}, \dots, v_{m})$ is linearly dependent and, in particular, is not a basis.

Note (The coordinate map is linear)

The coordinate map

{[\cdot]}_{ℬ} : V ⟶ 𝕂^{n}

is linear, since

\begin{matrix} {[α v]}_{ℬ} = α {[v]}_{ℬ} \\ {[u + v]}_{ℬ} = {[u]}_{ℬ} + {[v]}_{ℬ} \end{matrix}} ⟹ {[\cdot]}_{ℬ} is linear.

Proof of Lemma 1

Consider

x_{1} v_{1} + \dots + x_{m} v_{m} = 0.

Taking coordinates with respect to $ℬ$ gives

x_{1} {[v_{1}]}_{ℬ} + \dots + x_{m} {[v_{m}]}_{ℬ} = 0.

The zero vector here has $n$ components. Thus,

M [\begin{matrix} x_{1} \\ ⋮ \\ x_{m} \end{matrix}] = 0, M = ([v_{1}]_{ℬ}, \dots, [v_{m}]_{ℬ}) \in 𝕂^{n \times m} .

If $m > n$ , this system has infinitely many solutions in $𝕂^{m}$ . Consequently, $(v_{1}, \dots, v_{m})$ is linearly dependent. $□$

Proof of Theorem 1, statement 3

Let $A$ and $B$ be two bases of $V$ . We want to show that $| A | = | B |$ .

First, $| A | \leq | B |$ . Otherwise, if $| A | > | B | = n$ , Lemma 1 implies that $A$ is linearly dependent, so $A$ is not a basis, which is a contradiction.

The inequality $| B | \leq | A |$ follows analogously. Hence,

| A | = | B | . □

Linear transformations

Notation and terminology

A linear transformation is written as

T : V ⟶ W, v ⟼ T v .

Definition (Null space)

The null space, or kernel, of $T$ is

Null (T) = N (T) = \ker T = {v \in V : T v = 0} \subseteq V .

Definition (Image)

The image of $T$ is

Ran (T) = Im (T) = T (V) = {T v : v \in V} \subseteq W .

Theorem (Rank–nullity theorem)

For a linear transformation $T : V \to W$ ,

\dim V = \dim (\ker T) + rank (T),

where

rank (T) = \dim (Ran (T)) .

Matrix representation of a linear transformation

Let

T : V ⟶ W,

where $ℬ = (b_{1}, \dots, b_{m})$ is a basis of $V$ and $𝒞 = (c_{1}, \dots, c_{n})$ is a basis of $W$ .

The corresponding coordinate maps are

{[\cdot]}_{ℬ} : V ⟶ 𝕂^{m}

and

{[\cdot]}_{𝒞} : W ⟶ 𝕂^{n} .

Thus, in coordinates, the linear transformation can be represented as

𝕂^{m} ⟶ 𝕂^{n} .

Definition (Matrix representation)

The matrix representation of $T$ with respect to the bases $ℬ$ and $𝒞$ is denoted by

{[T]}_{𝒞 ℬ} \in 𝕂^{n \times m} .

It satisfies

{[T v]}_{𝒞} = {[T]}_{𝒞 ℬ} {[v]}_{ℬ} .

Note that the bases are written from right to left: $ℬ$ is the basis of the domain $V$ , while $𝒞$ is the basis of the codomain $W$ . Therefore, we write

{[T]}_{𝒞 ℬ}

rather than

{[T]}_{ℬ 𝒞} .