Basis change as a natural isomorphism

The aim of this blog is to elucidate why basis change constitutes a natural transformation within the context of vector spaces.

Consider the categories of finite-dimensional vector spaces over a field $\mathbb{F}$$\mathbb{F}$, and that of ${\mathbb{F}}^n$$\mathbb{F}^n$. There exists a functor ${\mathcal{B}}_i:\mathcal{C}\to \mathcal{D}$$\mathcal{B}_i: \mathcal{C} \to \mathcal{D}$ linking these categories, which depends on the basis system selected for each vector space.

If $\dim V=n$$\dim V = n$, then ${\mathcal{B}}_i(V)={\mathbb{F}}^n$$\mathcal{B}_i(V) = \mathbb{F}^n$. For morphisms, it is well-known that ${\mathcal{B}}_i(T)$$\mathcal{B}_i(T)$ represents the matrix form of a linear map.

Let's explore a linear map:

$$\begin{array}{}\text{(1)}& T:V\to W\end{array}$$

Suppose $v_1,\dots ,v_n$$v_1, \ldots, v_n$ are the basis vectors of $V$$V$ and $w^1,\dots ,w^m$$w^1, \ldots, w^m$ form the dual basis with respect to the chosen basis of $W$$W$.

The matrix representation of $T$$T$ is then given by ${\mathcal{B}}_i(T):T⟼(a_{i,j})$$\mathcal{B}_i(T): T \longmapsto (a_{i,j})$, where $a_{i,j}=w^i(T{v}_j)$$a_{i,j} = w^i(Tv_j)$.

For more details and an illustrative example, please refer to my previous blog post.

Now, consider a different basis system and the associated functor ${\mathcal{B}}_j$$\mathcal{B}_j$. The transformation between these bases is a natural isomorphism, illustrated by the following commutative diagram:

$$\begin{array}{}\text{(2)}& \begin{array}{ccc}{\mathcal{B}}_i(V)& \stackrel{{\mathcal{B}}_i(T)}{\to }& {\mathcal{B}}_i(W)\\ {\eta }_V\downarrow & & \downarrow {\eta }_W& \\ {\mathcal{B}}_j(V)& \underset{{\mathcal{B}}_j(T)}{\overset{}{\to }}& {\mathcal{B}}_j(W)\end{array}\end{array}$$

Here, $\eta$$\eta$ is defined as:

$$\begin{array}{}\text{(3)}& {\mathcal{B}}_j\circ {\mathcal{B}}_i^{-1}\end{array}$$

Since the functors ${\mathcal{B}}_i$$\mathcal{B}_i$ and ${\mathcal{B}}_j$$\mathcal{B}_j$ each provide a family of isomorphism $S_V^i:V\to {\mathbb{F}}^n$$S^i_V: V \to \mathbb{F}^n$.

Hence

$$\begin{array}{}\text{(4)}& {\eta }_V:=S_V^j\circ (S_V^i{)}^{-1}\end{array}$$

Indeed, each $\eta$$\eta$ represents the unique linear map that transforms one basis to another.

For instance, if $v_1,\dots ,v_n$$v_1, \ldots, v_n$ and $u_1,\dots ,u_n$$u_1, \ldots, u_n$ represent two different chosen bases of $V$$V$, then ${\eta }_V$$\eta_V$ is the unique linear map satisfies that

$$\begin{array}{}\text{(5)}& \mathrm{\forall }i\in \{1,...,n\},{\eta }_V(v_i)=u_i\end{array}$$

This treatment shows how basis change in vector spaces can be seen as a natural transformation, crucial for understanding the interplay between different coordinate systems and their equivalence under linear transformations.

Remark. I think it is clear enough that given a family of vectors in $V$$V$ such as $v_1,...,v_k$$v_1,...,v_k$ it determined a unique linear transformation ${\mathrm{Span}}_{v_1,...,v_k}:{\mathbb{F}}^k\to V$$\mathrm{Span}_{v_1,...,v_k}:\mathbb F^k\to V$. When ${\mathrm{Span}}_{v_1,...,v_k}$$\mathrm{Span}_{v_1,...,v_k}$ is injective iff $v_1,...,v_n$$v_1,...,v_n$ is linearly independent, and ${\mathrm{Span}}_{v_1,...,v_k}$$\mathrm{Span}_{v_1,...,v_k}$ is an isomorphism iff $v_1,...,v_k$$v_1,...,v_k$ is a basis of $V$$V$.

Each $S_V^i$$S^i_V$ is the inverse of Span respect to the basis you selecte.

We are in particular interested in the case that $V=W={\mathbb{F}}^n$$V=W=\mathbb F^n$. If $v_1,...,v_n$$v_1,...,v_n$ and $w_1,...,w_n$$w_1,...,w_n$ is the basis you selecte.

Since ${\eta }_V(v_i)=w_i$$\eta_V(v_i)=w_i$, hence the natural isomorphism is given b $P=[w_1,...,w_n][v_1,...,v_n{]}^{-1}$$P=[w_1,...,w_n][v_1,...,v_n]^{-1}$.