Vector Spaces and Linear Maps

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, so it has a basis, a well-defined dimension, and an isomorphism with a coordinate space $F^n$ — the apparatus the canonical forms run on.¹

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

Bases and dimension

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

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

Every independent set is therefore no larger than every basis, and every spanning set no smaller, so any two bases have the same size.²

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

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 $F[x]$.

Dimension counts across quotients and maps

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

The dimension of $\ker \phi$ is the nullity, that of $\phi (V)$ 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 ($2\mathbb{Z}\cong \mathbb{Z}$).

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 $\phi :V\to W$ is determined by its values on $\mathcal{B}$, and writing $\phi (v_j)={\sum }_i{a}_{ij}{w}_i$ collects those values into a matrix.

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

Taking $V=W$ and one basis, ${\mathrm{End}}_F(V)\cong M_n(F)$ 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.³

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.

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 $\phi$,

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 ${\mathrm{End}}_F(V)$. 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.

Dual spaces

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.

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

Because the definition of $\phi$ never mentions a basis, $V$ and $V^{\ast \ast }$ are identified without choices, whereas $V\cong V^{\ast }$ requires picking one. A linear map $\phi :V\to W$ induces the transpose ${\phi }^{\ast }:W^{\ast }\to V^{\ast }$, $f\mapsto f\circ \phi$, whose matrix in the dual bases is the transpose of $M(\phi )$. 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 $W^{\ast }$.⁴

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.

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.⁵

• 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 $P^{-1}AP$ 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.

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 and reads off invariant factors, which produces the rational and Jordan canonical forms.