Ahhhhh it’s almost 2016 (only *three* more days)! I’m incredibly excited, mainly because I’ll be going to college this fall. In this blog post, I wanted to point out something which Noam Elkies informed me about, because I hadn’t recognized it and found it pretty awesome!

As the title suggests, we’ll take a look at the general linear group over a finite field. First, let be a prime, and let . Then the equation always has a unique solution. More generally, if , always has a *unique *solution if . Thus .

What about ? If , then a necessary and sufficient condition for to be invertible is that . We *could* proceed via a case-by-case analysis, but that is more difficult. Alternatively, we could simply note a matrix’s determinant is zero if and only if the columns are linearly independent that the vector must be nonzero (which gives us choices), and can’t be a multiple of , which gives us choices. Thus is .

In general, is . We can use basic group theory to determine as well. Note that is the kernel of the determinant function . In particular, using the isomorphism theorems, , so rearranging gives .

This is all fun stuff, but what we’re really here for is the star of the day (or rather, year): . Using the above formula, we get: . Excluding the trivial cases (i.e. ), we’re not going to have this for another 1512 years! So enjoy 2016, and, as always, have fun!

Happy ,

SKD

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