---
title: Vector Spaces and Linear Maps
module: Module Theory
moduleNumber: 8
lessonNumber: 4
order: 804
summary: >
  A vector space is a module over a field, and the field hypothesis removes every
  pathology a general module can have: every vector space is free, so it has a
  basis, a well-defined dimension, and a coordinate isomorphism with F^n. Linear
  maps become matrices, change of basis becomes similarity, every space pairs with
  a dual of the same dimension, and the determinant is the unique alternating
  multilinear normalized form.
topics: [Module Theory]
draft: false
sources:
  - book: Dummit & Foote
    ref: "Ch. 11 — Vector Spaces; §11.1 Definitions and Basic Theory, §11.2 The Matrix of a Linear Transformation"
  - book: Dummit & Foote
    ref: "§11.3 Dual Vector Spaces; §11.4 Determinants"
---

A **vector space** over a field $F$ is exactly an $F$-module: the module axioms,
with the ring a field, are the vector-space axioms verbatim. A general module can
lack a basis, carry torsion, and have no coordinate description; over a field none
of that occurs. Every vector space is
[free](/abstract-algebra/module-theory/free-modules-and-direct-sums), so it has a
basis, a well-defined dimension, and an isomorphism with a coordinate space $F^n$ —
the apparatus the
[canonical forms](/abstract-algebra/modules-over-pids/rational-canonical-form)
run on.[^df-vs]

The terminology shifts when $R = F$ is a field, though the definitions do not.

| Module language | Vector-space language |
| --- | --- |
| $R$-module $M$ | vector space $M$ over $F$ |
| element $m$; scalar $a \in R$ | vector $m$; scalar $a \in F$ |
| submodule; quotient module | subspace; quotient space |
| free module of rank $n$ | space of dimension $n$ |
| module homomorphism | linear transformation |
| $M = RA$ (generated by $A$) | $M = \operatorname{Span}(A)$ |

## Bases and dimension

> **Definition (Independence, basis).** A set $S \subseteq V$ is **linearly
> independent** if $a_1 v_1 + \cdots + a_n v_n = 0$ with $v_i \in S$ forces all
> $a_i = 0$. A **basis** is an ordered, linearly independent set that spans $V$.

The first structural fact is that finitely generated vector spaces are free.

> **Proposition.** If $A = \{v_1, \dots, v_n\}$ spans $V$ but no proper subset spans
> $V$, then $A$ is a basis. In particular, every finitely generated vector space is a
> free $F$-module, and every finite spanning set contains a basis.

> **Proof.** If $A$ were dependent, some $v_i$ would be a combination of the
> others and could be deleted while still spanning, contradicting minimality.
> Removing dependent vectors from any spanning set thus lands on a basis. $\square$

Dimension rests on one further fact, the exchange of one spanning set for another.

> **Theorem (Replacement).** If $A = \{a_1, \dots, a_n\}$ is a basis of $V$ and
> $\{b_1, \dots, b_m\}$ is linearly independent, then after reordering, each $b_k$ can
> replace an $a_k$ with $\{b_1, \dots, b_k, a_{k+1}, \dots, a_n\}$ still a basis. In
> particular $m \le n$.

Every independent set is therefore no larger than every basis, and every spanning set
no smaller, so any two bases have the same size.[^df-basis]

> **Definition (Dimension).** The common cardinality of the bases of $V$ is
> $\dim_F V$. If no finite basis exists, $\dim V = \infty$.

Two further consequences fall out. The **building-up lemma** says any independent set
in a finite-dimensional space extends to a basis. And choosing a basis $v_1, \dots,
v_n$ gives the **coordinate isomorphism**: the map $F^n \to V$ sending
$(a_1, \dots, a_n) \mapsto \sum a_i v_i$ is linear, surjective (the $v_i$ span), and
injective (they are independent), so

$$
V \;\cong\; F^n, \qquad n = \dim_F V.
$$

Any two spaces of the same dimension are isomorphic; dimension is a complete
invariant. Beyond finite dimension, Zorn's lemma still yields a basis, so **every**
vector space is free — a field's lack of proper nonzero ideals
removes the torsion obstruction that blocks a basis over $\mathbb{Z}$ or $F[x]$.

$$
% caption: Choosing a basis identifies an $n$-dimensional space with coordinate
% space: each vector is recorded by its unique tuple of scalar coordinates, and the
% correspondence is a linear isomorphism.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=30mm, minimum height=13mm, align=center, font=\footnotesize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[box, draw=acc, text=acc, thick] (V) at (0,0) {space $V$\\$v = a_1 v_1 +$ ... $+ a_n v_n$};
  \node[box] (F) at (6.4,0) {coordinate space $F^n$\\scalars $a_1$ to $a_n$};
  \draw[->, acc, thick] (V.10) -- (F.170) node[midway, above, font=\scriptsize, text=black] {coordinates};
  \draw[->, black] (F.190) -- (V.-10) node[midway, below, font=\scriptsize] {rebuild};
\end{tikzpicture}
$$

### Dimension counts across quotients and maps

> **Theorem (Dimension of a quotient).** For a subspace $W \le V$,
> $\dim V = \dim W + \dim(V/W)$.

> **Proof.** Extend a basis of $W$ to a basis of $V$; the added vectors project to
> a basis of $V/W$. $\square$

Specializing $W = \ker\varphi$ and $V/W \cong \varphi(V)$ gives the
rank–nullity theorem.

> **Corollary (Rank–nullity).** For a linear map $\varphi : V \to U$,
> $\dim V = \dim\ker\varphi + \dim\varphi(V)$.

The dimension of $\ker\varphi$ is the **nullity**, that of $\varphi(V)$ the **rank**.
For a map between spaces of equal finite dimension, injective, surjective, and
bijective all coincide — a counting statement with no analogue for general modules,
where a proper submodule can be isomorphic to the whole ($2\mathbb{Z} \cong
\mathbb{Z}$).

> **Worked example.** Let $\varphi : \mathbb{Q}^3 \to \mathbb{Q}^2$ be given by
> $\varphi(x, y, z) = (x + 2y + z,\; 2x + 4y + 3z)$. Compute the rank and nullity.
>
> The kernel is the solution set of $x + 2y + z = 0$ and $2x + 4y + 3z = 0$.
> Subtracting twice the first equation from the second leaves $z = 0$, and then
> $x = -2y$, so
>
> $$
> \ker\varphi = \{(-2y,\; y,\; 0) : y \in \mathbb{Q}\} = \operatorname{Span}\bigl((-2, 1, 0)\bigr),
> $$
>
> of dimension $1$ (nullity $1$). The image contains $\varphi(e_1) = (1, 2)$ and
> $\varphi(e_3) = (1, 3)$, which are independent, so the rank is $2$ and
> $\varphi$ is surjective. Rank–nullity checks: $3 = 1 + 2$.

## The matrix of a linear transformation

Fix ordered bases $\mathcal{B} = \{v_1, \dots, v_n\}$ of $V$ and $\mathcal{E} =
\{w_1, \dots, w_m\}$ of $W$. A linear map $\varphi : V \to W$ is determined by its
values on $\mathcal{B}$, and writing $\varphi(v_j) = \sum_i a_{ij} w_i$ collects those
values into a matrix.

> **Definition (Matrix of a map).** The **matrix of $\varphi$** with respect to
> $\mathcal{B}, \mathcal{E}$ is $M_{\mathcal{B}}^{\mathcal{E}}(\varphi) = (a_{ij}) \in
> M_{m \times n}(F)$, whose $j$-th column holds the $\mathcal{E}$-coordinates of
> $\varphi(v_j)$. Then $[\varphi(v)]_{\mathcal{E}} =
> M_{\mathcal{B}}^{\mathcal{E}}(\varphi)\,[v]_{\mathcal{B}}$.

This correspondence is an isomorphism of vector spaces, and it turns composition into
matrix multiplication.

> **Theorem (Representation).** The map $\varphi \mapsto
> M_{\mathcal{B}}^{\mathcal{E}}(\varphi)$ is an isomorphism $\operatorname{Hom}_F(V, W)
> \cong M_{m \times n}(F)$; hence $\dim\operatorname{Hom}_F(V, W) = (\dim V)(\dim W)$.
> For composable maps, $M(\varphi \circ \psi) = M(\varphi)\,M(\psi)$ over compatible
> bases.

Taking $V = W$ and one basis, $\operatorname{End}_F(V) \cong M_n(F)$ is a ring
isomorphism, and it restricts to the group isomorphism $\mathrm{GL}(V) \cong
\mathrm{GL}_n(F)$ between invertible operators and invertible matrices. Matrix
multiplication is associative because function composition is, without any direct
computation.[^df-matrix]

### Change of basis and similarity

A different choice of basis gives a different matrix for the _same_ operator, and the
two are related by conjugation.

> **Definition (Similarity).** Matrices $A, B \in M_n(F)$ are **similar** if
> $B = P^{-1} A P$ for some invertible $P$.

Let $\mathcal{B}, \mathcal{E}$ be two bases of $V$ and let $P =
M_{\mathcal{E}}^{\mathcal{B}}(I)$ be the **transition matrix**, whose columns express
the $\mathcal{E}$-vectors in $\mathcal{B}$-coordinates. Then for any operator
$\varphi$,

$$
M_{\mathcal{E}}^{\mathcal{E}}(\varphi) \;=\; P^{-1}\, M_{\mathcal{B}}^{\mathcal{B}}(\varphi)\, P.
$$

So the matrices of one operator in different bases form a single similarity class, and
conversely similar matrices represent one operator in different bases. Change of basis
_is_ conjugation in $\mathrm{GL}(V)$; the similarity classes are the orbits of
$\mathrm{GL}(V)$ acting by conjugation on $\operatorname{End}_F(V)$. Choosing the
basis that makes the matrix as simple as possible is the canonical-forms problem,
solved by viewing $V$ as an $F[x]$-module and applying the
[structure theorem over a PID](/abstract-algebra/modules-over-pids/structure-theorem-over-pids).

$$
% caption: One operator, two bases: reading it in basis $\mathcal{B}$ gives matrix
% $A$, in basis $\mathcal{E}$ gives $C$, and the transition matrix $P$ conjugates
% one into the other, $C = P^{-1} A P$.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=22mm, minimum height=11mm, align=center, font=\scriptsize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[box] (BB) at (0,2.4) {$V$ in basis $B$};
  \node[box] (BB2) at (4.4,2.4) {$V$ in basis $B$};
  \node[box] (EE) at (0,0) {$V$ in basis $E$};
  \node[box] (EE2) at (4.4,0) {$V$ in basis $E$};
  \draw[->, acc, thick] (BB) -- (BB2) node[midway, above, font=\scriptsize, text=black] {$A$};
  \draw[->, acc, thick] (EE) -- (EE2) node[midway, below, font=\scriptsize, text=black] {$C$};
  \draw[->, black] (EE) -- (BB) node[midway, left, font=\scriptsize] {$P$};
  \draw[->, black] (EE2) -- (BB2) node[midway, right, font=\scriptsize] {$P$};
\end{tikzpicture}
$$

> **Worked example.** Let $T$ act on $\mathbb{Q}^2$ with standard-basis matrix
> $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$. Find a basis in which the
> matrix of $T$ is diagonal.
>
> The eigenvalues are the diagonal entries $3$ and $2$ (the matrix is
> triangular). Eigenvectors: for $\lambda = 3$, $(A - 3I)v = 0$ gives
> $v_1 = (1, 0)$; for $\lambda = 2$, $(A - 2I)v = 0$ reads
> $v_x + v_y = 0$, giving $v_2 = (1, -1)$. Put the new basis in the columns of
> the transition matrix and invert:
>
> $$
> P = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix},
> \qquad
> P^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}.
> $$
>
> Conjugating, first $AP$, then $P^{-1}(AP)$:
>
> $$
> AP = \begin{pmatrix} 3 & 2 \\ 0 & -2 \end{pmatrix},
> \qquad
> P^{-1} A P = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}
> \begin{pmatrix} 3 & 2 \\ 0 & -2 \end{pmatrix}
> = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}.
> $$
>
> In the basis $\{v_1, v_2\}$ the operator is $\operatorname{diag}(3, 2)$. As a
> check without matrices:
> $T(v_2) = T(1, -1) = (3 \cdot 1 + 1 \cdot (-1),\; 2 \cdot (-1)) = (2, -2) = 2v_2$,
> and $T(v_1) = (3, 0) = 3v_1$.

## Dual spaces

> **Definition (Dual space).** The **dual** of $V$ is $V^{\ast} =
> \operatorname{Hom}_F(V, F)$, the space of **linear functionals** on $V$.

For a basis $\{v_1, \dots, v_n\}$, define $v_i^{\ast} \in V^{\ast}$ by
$v_i^{\ast}(v_j) = \delta_{ij}$ (one if $i = j$, else zero). These are the **dual
basis**, and they are independent and span, so $\dim V^{\ast} = \dim V$ in finite
dimension. The functional $v_i^{\ast}$ reads off the $i$-th coordinate of a vector.

$$
% caption: The dual basis pairs each basis vector with the functional returning its
% coordinate: the functional dual to $v_i$ evaluates to one on $v_i$ and to zero on
% every other basis vector, the identity pattern shown in the grid.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % column headers: basis vectors
  \node[font=\scriptsize] at (1.0,2.3) {$v_1$};
  \node[font=\scriptsize] at (1.8,2.3) {$v_2$};
  \node[font=\scriptsize] at (2.6,2.3) {$v_3$};
  % row headers: dual functionals f_i
  \node[font=\scriptsize] at (0,1.4) {$f_1$};
  \node[font=\scriptsize] at (0,0.7) {$f_2$};
  \node[font=\scriptsize] at (0,0.0) {$f_3$};
  % grid values
  \node[font=\scriptsize, text=acc] at (1.0,1.4) {$1$};
  \node[font=\scriptsize, text=black] at (1.8,1.4) {$0$};
  \node[font=\scriptsize, text=black] at (2.6,1.4) {$0$};
  \node[font=\scriptsize, text=black] at (1.0,0.7) {$0$};
  \node[font=\scriptsize, text=acc] at (1.8,0.7) {$1$};
  \node[font=\scriptsize, text=black] at (2.6,0.7) {$0$};
  \node[font=\scriptsize, text=black] at (1.0,0.0) {$0$};
  \node[font=\scriptsize, text=black] at (1.8,0.0) {$0$};
  \node[font=\scriptsize, text=acc] at (2.6,0.0) {$1$};
  \draw[black] (0.6,1.75) -- (3.0,1.75);
  \draw[black] (0.4,1.75) -- (0.4,-0.35);
  \node[font=\scriptsize, anchor=west] at (3.4,0.7) {$f_i$ reads coordinate $i$};
\end{tikzpicture}
$$

> **Worked example.** Find the dual basis of $v_1 = (1, 1)$, $v_2 = (1, -1)$ in
> $\mathbb{Q}^2$.
>
> A functional on $\mathbb{Q}^2$ has the form $f(x, y) = ax + by$. The condition
> $f_1(v_1) = 1$, $f_1(v_2) = 0$ is the linear system $a + b = 1$, $a - b = 0$,
> so $a = b = \tfrac{1}{2}$:
>
> $$
> f_1(x, y) = \tfrac{1}{2}(x + y).
> $$
>
> Similarly $f_2(v_1) = 0$, $f_2(v_2) = 1$ gives $a + b = 0$, $a - b = 1$, so
>
> $$
> f_2(x, y) = \tfrac{1}{2}(x - y).
> $$
>
> Check: $f_1(1,1) = 1$, $f_1(1,-1) = 0$, $f_2(1,1) = 0$, $f_2(1,-1) = 1$. Any
> vector reconstructs from its dual-basis readings:
> $(x, y) = f_1(x,y)\, v_1 + f_2(x,y)\, v_2$, e.g.
> $(3, 1) = 2 v_1 + 1 v_2$ with $f_1(3,1) = 2$, $f_2(3,1) = 1$.

The dual of the dual returns the original space, and — unlike the isomorphism $V
\cong F^n$ — the identification is canonical.

> **Theorem (Double dual).** The evaluation map $\varphi : V \to V^{\ast\ast}$, $v
> \mapsto (f \mapsto f(v))$, is an injective linear map, natural (basis-independent),
> and an isomorphism when $\dim V < \infty$.

Because the definition of $\varphi$ never mentions a basis, $V$ and $V^{\ast\ast}$ are
identified without choices, whereas $V \cong V^{\ast}$ requires picking one. A linear
map $\varphi : V \to W$ induces the **transpose** $\varphi^{\ast} : W^{\ast} \to
V^{\ast}$, $f \mapsto f \circ \varphi$, whose matrix in the dual bases is the
transpose of $M(\varphi)$. That fact yields a clean proof that the **row rank equals
the column rank** of any matrix: both equal the rank of the associated map, computed
on $V$ or on $W^{\ast}$.[^df-dual]

## Determinants

The determinant is the unique scalar built from the columns of a matrix that is
linear in each column, vanishes when two columns coincide, and normalizes the
identity to $1$. This works over any commutative ring $R$, not only a field.

> **Definition (Alternating multilinear form).** A map $\varphi : V^n \to R$ is
> **$n$-multilinear** if it is $R$-linear in each argument with the others fixed, and
> **alternating** if $\varphi(\dots, v_i, v_{i+1}, \dots) = 0$ whenever two adjacent
> arguments are equal. A **determinant function** is an alternating $n$-multilinear
> form on the columns with $\det(I) = 1$.

> **Theorem (Existence and formula).** There is a unique determinant function on
> $M_n(R)$, given by the Leibniz sum
> $$
> \det(a_{ij}) \;=\; \sum_{\sigma \in S_n} \varepsilon(\sigma)\,
> a_{\sigma(1)1}\, a_{\sigma(2)2} \cdots a_{\sigma(n)n},
> $$
> where $\varepsilon(\sigma)$ is the sign of the permutation.

The sign appears because swapping two arguments of an alternating form negates it, so
each permutation contributes with its parity. From the formula, several properties are
immediate and complete the calculus of determinants.[^df-det]

- **Transpose invariance.** $\det(A^{\top}) = \det(A)$, so the determinant is equally
  a form in the rows; row and column statements coincide.
- **Multiplicativity.** $\det(AB) = \det(A)\det(B)$. In particular a change of basis
  $P^{-1} A P$ leaves the determinant unchanged, so the **determinant of a linear
  operator** is well defined, independent of basis.
- **Singularity test.** Over an integral domain, $\det A = 0$ if and only if the
  columns (equivalently the rows) are linearly dependent; over a field, $A$ is
  invertible iff $\det A \ne 0$.
- **Cramer's rule and cofactors.** $\beta_i \det A = \det(A_1, \dots, B, \dots, A_n)$
  when $B = \sum \beta_i A_i$; the cofactor expansion computes $\det A$ along any row,
  and the transposed cofactor matrix gives $A^{-1}$ when $\det A$ is a unit.

Geometrically, over $\mathbb{R}$ the determinant is the signed volume scaling of the
unit cube: a linear map sends the cube spanned by the standard basis to the
parallelepiped spanned by the columns of its matrix, and $|\det A|$ is that region's
volume, with the sign recording orientation.

$$
% caption: Over the reals the determinant is the signed area (in higher dimension,
% volume) scaling factor: the unit square spanned by the basis maps to the
% parallelogram spanned by the columns of $A$, whose area is the absolute
% determinant.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % unit square
  \fill[black] (0,0) rectangle (1.4,1.4);
  \draw[black] (0,0) rectangle (1.4,1.4);
  \node[font=\scriptsize] at (0.7,0.7) {area $1$};
  \draw[->, black] (0,0) -- (1.4,0);
  \draw[->, black] (0,0) -- (0,1.4);
  % arrow
  \draw[->, acc, very thick] (2.2,0.7) -- (3.6,0.7) node[midway, above, font=\scriptsize, text=black] {$A$};
  % image parallelogram
  \begin{scope}[xshift=4.2cm]
    \fill[acc!14] (0,0) -- (2.1,0.6) -- (2.8,2.4) -- (0.7,1.8) -- cycle;
    \draw[acc] (0,0) -- (2.1,0.6) -- (2.8,2.4) -- (0.7,1.8) -- cycle;
    \draw[->, acc, thick] (0,0) -- (2.1,0.6);
    \draw[->, acc, thick] (0,0) -- (0.7,1.8);
    \node[font=\scriptsize, text=acc] at (1.5,1.0) {area $= \det A$};
  \end{scope}
\end{tikzpicture}
$$

## General modules versus vector spaces

Because a vector space is free with a basis, an operator becomes a matrix, and the
remaining question is which basis makes that matrix simplest. Answering it views
$(V, T)$ as an
[$F[x]$-module](/abstract-algebra/module-theory/introduction-to-modules) and reads
off invariant factors, which produces the
[rational](/abstract-algebra/modules-over-pids/rational-canonical-form) and
[Jordan](/abstract-algebra/modules-over-pids/jordan-canonical-form) canonical
forms.

| Property | General $R$-module | Vector space over $F$ |
| --- | --- | --- |
| Basis exists | not always | always (Zorn) |
| Complete invariant | invariant factors | dimension |
| Submodule $\cong$ whole possible | yes ($2\mathbb{Z} \cong \mathbb{Z}$) | no (dimension counts) |
| Every short exact sequence splits | no | yes |
| $V \cong V^{\ast\ast}$ | not in general | canonically (finite dim) |

[^df-vs]: **Dummit & Foote**, _Abstract Algebra_, 3rd ed., §11.1 — vector spaces as modules over a field, the terminology dictionary, and the proof that finitely generated (and, via Zorn's lemma, arbitrary) vector spaces are free.
[^df-basis]: **Dummit & Foote**, §11.1, Theorem 3 (Replacement) and Corollaries 4–8 — invariance of basis size, dimension, the building-up lemma, $V \cong F^n$, and $\dim V = \dim W + \dim(V/W)$ with rank–nullity.
[^df-matrix]: **Dummit & Foote**, §11.2, Theorems 10–14 — the matrix of a linear transformation, $\operatorname{Hom}_F(V, W) \cong M_{m \times n}(F)$, matrices of composites, $\operatorname{End}_F(V) \cong M_n(F)$, and change of basis by similarity through the transition matrix.
[^df-dual]: **Dummit & Foote**, §11.3, Proposition 18 and Theorems 19–20, Corollary 21 — the dual basis, the natural isomorphism $V \cong V^{\ast\ast}$, the transpose of a linear map, and the equality of row and column rank.
[^df-det]: **Dummit & Foote**, §11.4, Theorem 24 and Corollaries 25–27, Theorems 28–30 — the determinant as the unique alternating multilinear normalized form, the Leibniz formula, transpose invariance, multiplicativity, the singularity criterion, Cramer's rule, and the cofactor formula for the inverse.