# How can a vector dual space be used to model division?

A couple of months ago I posted a question about how to formally define systems of units, and someone posted this fascinating response, explaining that each base unit can be the basis of a single dimensional vector space and we can then use the tensor product to model multiplication of base units to form composite units. That part makes sense.

The part I don't entirely understand is that they say we can use the dual space to model division of base units. e.g. If $$M$$ is the vector space for meters and $$S$$ for seconds, then the composite unit of $$\frac{m}{s}$$ can be modeled as $$M \otimes S^*$$. How does that work exactly?

A dual space, $$V^*$$, is just the vector space of all linear functions that map $$V$$ to $$F$$, yes and division isn't a linear operation, so why should there be a member of the dual space corresponding to the reciprocal function?

On a side note, I did post this question as a comment on the original post and the person responded by saying

An element $$α∈V^∗$$ can be multiplied with an element of $$v∈V$$ by means of $$α⋅v=α(v)$$. Now since $$V$$ is one-dimensional, every $$v∈V$$ is of the form $$v=λe$$ for some chosen non-zero $$e∈V$$. Now, define $$v^{−1}∈V^∗$$ by $$v^{−1}(μe)=μ⋅\frac{1}{λ}$$, which is clearly linear. Then $$v^{−1}⋅v=v^{−1}(λe)=\frac{λ}{λ}=1$$, which justifies the notation $$v^{−1}$$. Also, clearly $$v^{−1}$$ is the unique element of $$V^∗$$ with this property.

I still don't quite understand though. I assume $$\lambda$$ is meant to be a scalar, in which case, it seems like all that's actually happening is we're defining a linear function that simply multiplies by the reciprocal of a scalar. In particular, I don't follow how $$v^{−1}⋅v=v^{−1}(λe)=\frac{λ}{λ}=1$$. It seems like just the $$\lambda$$s would cancel out and we'd be left with $$e$$, not $$1$$.

• No, that’s the point. The basic linear map acts on $e$ and returns $1$. (By the basic linear map I mean the basis $\{e^*\}$ for $V^*$ dual to the basis $\{e\}$ for $V$.) May 27 at 23:21