# The relationship between vector space dualization and matrix transposition.

Let $$\mathbb{F}$$ be a field. The category of matrices $$\mathbf{Mat}$$ has $$\mathbb{N}_0$$ as class of objects and $$\mathrm{hom}(n,m)=\mathbb{F}^{n\times m}$$ for $$n,m\in\mathbb{N}_0$$, with $$\mathrm{id}_n=\mathbf{1}_n$$ and $$B\circ A=A\cdot B$$ with the usual matrix multiplication whenever defined. The category of finite-dimensional vector spaces $$\mathbf{FinVect}$$ has $$\mathbb{F}$$-vector spaces as objects and linear maps as morphisms, where composition is the usual composition of functions and the identity morphism is the identity map. The categories $$\mathbf{Mat}$$ and $$\mathbf{FinVect}$$ are equivalent.

There is a contravariant duality functor $$(-)^\ast\colon\mathbf{FinVect}\rightarrow\mathbf{FinVect}$$ with $$V^\ast=\mathrm{Hom}(V,\mathbb{F})$$ for any vector space $$V$$ and $$f^\ast\colon W^\ast\rightarrow V^\ast,\,g\mapsto g\circ f$$ for any linear map $$f\colon V\rightarrow W$$. There also is a contravariant transpose functor $$(-)^t\colon\mathbf{Mat}\rightarrow\mathbf{Mat}$$ where $$n^t=n$$ for $$n\in\mathbb{N}_0$$ and $$A^t$$ is the usual transpose of a matrix.

These two contravariant functors on equivalent categories are related by the fact that for finite-dimensional vector spaces $$V,W$$ with bases $$\mathscr{B},\mathscr{C}$$ respectively, we have that $$(T^{\mathscr{B}}_{\mathscr{C}}(f))^t=T^{\mathscr{C}^\ast}_{\mathscr{B}^\ast}(f^\ast),$$ where $$T$$ denotes the transformation matrix for the respective bases.

Is there a way to describe this relationship of functors in category-theoretical terms?

There's an obvious functor $$\text{Mat} \to \text{FinVect}$$ which is an equivalence, but no obvious choice of inverse to this functor; choosing an inverse amounts to choosing, for each finite-dimensional vector space (note that there is more than a set's worth of such things!), a basis of that vector space. Not a fun operation to perform; we need something like the axiom of global choice to do it.
What can be done without making such choices is the following: there is an intermediate category which I'll just call $$\text{Base}$$, the category of finite-dimensional vector spaces together with a choice of basis (morphisms are still just linear transformations). There is an obvious functor $$F : \text{Base} \to \text{FinVect}$$ given by forgetting the basis, and an obvious functor $$M : \text{Base} \to \text{Mat}$$ given by writing linear transformations with respect to a basis. Both of these functors are equivalences, but again the functor $$F : \text{Base} \to \text{FinVect}$$ can't really be inverted obviously. See also anafunctor.
$$\text{Base}$$ has a dualization endofunctor $$D : \text{Base} \to \text{Base}$$ which takes duals of linear transformations and sends a pair $$(V, B)$$ of a vector space and a basis to the pair $$(V^{\ast}, B^{\ast})$$ of the dual vector space and the dual basis. Now what we can say is that the functors $$F$$ and $$M$$ intertwine this dualization functor and the other dualization functors, in the sense that we have not just natural isomorphisms but equalities
$$F \circ D = (-)^{\ast} \circ F : \text{Base} \to \text{FinVect}$$ $$M \circ D = (-)^T \circ M : \text{Base} \to \text{Mat}.$$