# Yoneda lemma applied to finite dimensional vector spaces.

This question is stenmed from my attempts to review my underestimating of the basic theory of finite dimensional vector spaces in a categorical language, and particularly to highlight he functorial behavior of the concept of base. My hope is to see possible corrections, clarifications or link to a more through expansions of such attempts:

Background:

A vector space is finite dimensional if it has a finite spanning subset. Let us denote the category of finite dimensional vector spaces and linear transformations by $\mathcal{V}$ and the well known category of matrices by $\mathcal{M}$, assuming that the underlying field is understood. Recall that the object of $\mathcal{M}$ are natural numbers and morphisms from $m$ to $n$ are matrices with $m$ rows and $n$ columns. (Fortunately, the matrix multiplication and the famous identity $I^{n\times n}$ matrices composition satisfy the axioms of category).

Question:

Obviously, a base has a functorial nature but how one should introduce the concept of base, purely in categorical language? Is the base a representable functor and does the canonical way of obtaining a (standard) base has anything to do with Yonada lemma?

Thanks.

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I'm reasonably confident there's nothing that will turn out to be canonical, in the categorical sense, about the so-called "standard" basis, which is just a notational fiction. – Kevin Carlson Nov 12 '12 at 23:48
What do you mean by that a base has a functorial nature? – Makoto Kato Nov 12 '12 at 23:52
It is just restating that base provides a coordination system. Once we are granted a base, $b\colon \mathcal{V}\rightarrow\mathcal{M}$, any vector spaces can be mapped to a column matrix and any linear transformation to a matrix. – Hooman Nov 13 '12 at 0:00

There is a distinguished $1$-dimensional vector space called $k$, the base field. This $1$-dimensional vector space is distinguished by the fact that it is equipped with canonical isomorphisms

$$k \otimes V \cong V \cong V \otimes k$$

satisfying certain identities making $k$ the monoidal unit for the tensor product. These canonical isomorphisms, among other things, distinguish the basis vector $1 \in k$. The reason is that the above isomorphisms specify actions of $k$ on each finite-dimensional vector space $V$ and $1 \in k$ is uniquely specified as the element of $k$ which acts by the identity.

A basis of a finite-dimensional vector space $V$ is a choice of isomorphism

$$k \oplus ... \oplus k \to V.$$

Here it is crucial that we use $k$ (equipped with the maps above) and not just a $1$-dimensional vector space. Beyond this it is not clear to me what you mean by "a base has a functorial nature."

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I like to think of an (ordered) basis $B=\langle b_1,\dots,b_n\rangle$ for an $n$-dimensional vector space $V$ as an isomorphism $\beta:F^n\to V$ (where $F$ is the field of scalars), namely the isomorphism sending the standard basis vectors $e_i=(0,\dots,0,1,0,\dots,0)$ (with 1 in the $i$-th position) to the corresponding $b_i$.

From this point of view, $\beta^{-1}$ codifies how $B$ allows us to represent a vector in $V$ by an $n$-tuple of numbers. The matrix for transforming from one basis to another is given by $\beta^{-1}\beta'$ (where $\beta$ and $\beta'$ are the two bases viewed as isomorphisms).

Note that the full subcategory of $\mathcal V$ consisting of the objects $F^n$ is isomorphic to your category $\mathcal M$ of matrices, and it is a skeleton of $\mathcal V$. In particular, the two categories are equivalent. To choose bases for all vector spaces in $\mathcal V$ is just to choose a particular equivalence between them.

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