The size of a finite skeletal category C [in the sense of Leinster](http://arxiv.org/abs/math.CT/0610260) is defined as follows: Label the objects of C by integers 1,2,...,n and let aij be the number of morphisms from i to j (for i and j between 1 and n). The **size (or Euler characteristic) of C** is defined as the sum of the entries of the inverse of the nxn matrix A=(aij), if the inverse exists. 


Let Fq be a finite field with q elements. For every natural number i, there is up to isomorphism exactly one Fq-vector space Vi of dimension i. The number of linear maps from Vi to Vj is equal to qij. We ignore the zero dimensional vector space V0. Consider the infinite matrix


Q=(qij)


where rows and and columns are indexed by positive integers 1,2,3,... From now on let us treat q as a formal parameter, don't care about convergence issues, and set v=q-1.


**Is there a notion of an inverse of Q?** (The entries will probably be formal power series in v.) **If the answer is yes, what is a closed form for the sum of the entries of the inverse (as a formal power series in v), i.e. the size of the category of finite dimensional Fq-vector spaces?**


At least every truncation Qn of Q to an upper left nxn corner has an inverse for every positive integer n, since Qn is a Vandermonde matrix. What is the limit of the sum of the entries of Qn-1 as n goes to infinity? I believe the answer is a power series in v. Is there an explicit form?


How can you interpret the answer? Is it the Euler characteristic of some moduli space? Is it equal or related to a sum of 1/Gl(Vi)? Does something interesting happen at q=1?



What is the size of the category of finite dimensional F_q vector spaces? 
Following the observations made in the comments one can compute the sum of the entries of Qn-1. It turns out that Kevin Costello's formula is true for every n.


Let (a1, a2, ..., an) be the the transpose of the kth column vector of Qn-1. (Of course, this vector depends on k, but we omit the index k.) Qiaochu Yuan suggested to consider the polynomial


A(x) = a1x + a2x2 + ... + anxn.


The degree of A is n, therefore A is determined by values at n+1 points. But we know that A(0)=0 and that


A(qi) = deltaik for i = 1, 2, ... , n. 


By Lagrange interpolation, A(x) is equal to


x(x-q)(x-q2)...(x-qk-1)(x-qk+1)...(x-qn) / qk(qk-q)(qk-q2)...(qk-qk-1)(qk-qk+1)...(qk-qn). 


The sum a1+a2+...+an is equal to A(1). Let us work with quantized integers. We use the notation [k] = (1-qk)/(1-q) = 1+q+...+qk-1. (Note that people from quantum groups sometimes use a different convention.) Furthermore, let [n choose k] be the quantized binomial coefficient. Then, A(1) is equal to 


(-1)k-1 [n choose k] qk(k-1)/2-kn.


We sum A(1) over all k. A variant of the [quantum binomial theorem](http://en.wikipedia.org/wiki/Gaussian_binomial) gives that the sum of the entries of Qn-1 is equal to 1 - (1-1/q)(1-1/q2)...(1-1/qn).


