Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:


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).


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?


share|cite|improve this question
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
up vote 7 down vote accepted

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."

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.