# how many invertible square matrices over a q-field

(practising for combinatorics exam) question starts with Euler's pentagonal numbers theorem. then goes onto asking - how many invertible nxn matrices over a field of q elements are there? what is the probability p(n,q) that a random nxn matrix is invertible?

• This is a nice question that I also usually give when teaching linear algebra. You can observe that the matrix is invertible if and only if all columns are linearly independent, and then imagine choosing columns one-by-one. There are $q^n-1$ choices of the first column (it only needs to be nonzero). Given this, there are...choices of the second column, and so on. Can you finish the problem? [I do not know what Euler's pentagonal number theorem is, but you can solve without it.] – Michael May 6 '17 at 20:26
• so for the second row it's q^(n)-q because we don't want a scalar multiple of row 1, etc until the last row where it's q^(n) - q^(n-1). then we multiply all these together ? – ananas May 6 '17 at 20:35
• Looks good. The probability question uses essentially the same reasoning. One aspect is you might want to formally prove that all $k$-dimensional subspaces of this vector space (with this $q$-field) have size $q^k$. If you like, you can answer your own question and mark it as "best answer" (this is standard practice on stackexchange when solving a question based on suggestions in the comments). – Michael May 6 '17 at 20:36
• A part of this question has been asked before : math.stackexchange.com/questions/1399406/… – Arnaud D. May 6 '17 at 21:44

## 1 Answer

This is the same as choosing a basis of the vector space $F^n$ where $F$ has q elements. So for the first basis element we have $q^n-1$ choices. For the next we have $q^n-q$. In general for the ith basis element we have $q^n-q^{i-1}$ choices. So the number of invertible matrices is the product of all these expressions from i= 1 to n