# Define dimension without referring to bases

The dimension of a vector space is the common cardinality of all bases. I would like to define it in a way that does not refer to bases. I only care about finite dimensional vector spaces.

First, I need to define what it means for a vector space to be finite dimensional without referring to a basis. One definition that works is to say that $$V$$ is finite-dimensional if the natural map $$V\rightarrow (V^*)^*$$ is an isomorphism (that's the double dual). I am open to other basis-free definitions.

As for defining what $$\dim V$$ is, my first thought was to define it as $$\text{trace}(I)$$. This is true, but it is not quite basis-free:

We have an isomorphism $$V^*\otimes V \cong \hom(V,V)$$. On $$V^*\otimes V$$, I can define the trace in a basis-free way. I can also define this map $$V^*\otimes V \rightarrow \hom(V,V)$$ without referring to a basis. However, I don't know how to show that it is an isomorphism without referring to bases. It would be especially nice if we could pinpoint of the identity map $$V\rightarrow V$$ as an element of $$V^*\otimes V$$ without referring to bases. Maybe it's the unique fixed point of some action or something like that?

Regarding the answers: At the moment there are three answers by Andreas Blass, Thorgott and me. I accepted one because I can only accept one. All three are worth reading in my opinion.

• Are you asking about vector spaces over a particular field $\mathbb F$? In that case we could ask whether $V$ is isomorphic to $\mathbb F^n$. – hardmath Apr 17 '19 at 18:01
• @hardmath: I am not sure I understand your question, but yes, I'm okay with fixing the field. I've just added an answer to my own question. – American Igor Apr 17 '19 at 18:12
• @hardmath: Oh, I get your point. Yeah, $\dim V$ is $n$ if $V$ is isomorphic to $\mathbb{F}^n$. But from this, I don't know how to see that $\dim V$ is well-defined without referring to bases. That is, a priori, maybe $\mathbb{F}^n$ and $\mathbb{F}^m$ are isomorphic for $n\neq m$. – American Igor Apr 17 '19 at 18:23

$$\dim(V)\colon=\sup\{n\in\mathbb{N}_0\vert\exists U_0\subsetneq U_1\subsetneq...\subsetneq U_n,\text{ subspaces of }V\}\in\mathbb{N}_0\cup\{\infty\}$$

I got it!

Let $$V$$ be a vector space over a field $$F$$. Assume that the natural map $$V\rightarrow V^{**}$$ given by $$v\mapsto(\varphi\mapsto \varphi(v))$$ is an isomorphism. Then, the natural map $$f:V\otimes V^*\rightarrow (V^*\otimes V)^*$$, given by $$f(v\otimes\varphi)=(\alpha\otimes u\mapsto\alpha(v)\cdot \varphi(u))$$, is an isomorphism. Consider $$\text{trace}\in (V^*\otimes V)^*$$, and define $$I=f^{-1}(\text{trace})$$. Define the dimension of $$V$$ to be $$\text{trace}(I)$$.

So the $$\dim V$$ is the "trace of trace"!

• The inconvenient of the trace definition is that it does not work in non-zero characteristic. Maybe it can be modified with some sort of functoriality, for instance to reduce to Witt vectors or something. – Captain Lama Apr 17 '19 at 19:18

Two options that are close to using bases but don't actually mention them:

(1) The dimension of $$V$$ is the smallest number of vectors that span $$V$$.

(2) The dimension of $$V$$ is the largest size of any linearly independent set of vectors in $$V$$.

To put it another way, the notion of basis combines the two notions of spanning and linear independence. But either one of these two notions is enough to define dimension; you don't need both.

I think the following works and is of a same flavour as the suggestion of Thorgott

Let $$P: V \to V$$ be a linear map. $$P$$ is said to be a projection map if $$P^2=P$$. Furthermore, we call a sequence of projection maps $$P_{1}, \dots, P_{n}$$ proper if $$P_{i}[im P_{i-1}] \neq im P_{i-1}$$ for all $$i \in \{2,\dots, n\}$$ and we say that a sequence of projection maps terminates at step $$i$$ if $$P_{i} \circ \dots \circ P_{1} = 0$$.

Now we call a vector space $$V$$ finite dimensional of dimension $$n$$ if there exists some proper sequence of projection maps of length $$P_{1}, \dots , P_{n}$$ such that it terminates