# Prove that Gl(n, F) is a group

I've seen "proofs" of this before but they all just assume we already know that matrix multiplication is associative, matrices with nonzero determinant are invertible, etc. and I do know this insofar that I've been told it since high school, but I don't think I've actually ever seen a proof that matrix multiplication is well defined, associative, etc. Does anyone know of any more rigorous proofs out there? I've googled a lot but haven't found anything. It's not exactly the sort of thing I'd want to spend time on proving rigorously myself since I already know these things to be true, but I would like to see a proof.

Thanks!

• Much of the proof comes down to the associativity of addition and multiplication and the well-definedness of these operations on the underlying field. I seem to remember a proof of some of these properties in Axler's Linear Algebra Done Right. – Matt Jun 7 '17 at 14:43
• Makes sense. I tend to be more algebra minded than analytic, but matrices and linear algebra for some reason don't seem to stick with me as well as other algebra. I'm reviewing to try to learn some representation theory of finite groups :) I'll see if there's an online PDF. Thank you! – Liam Cooney Jun 7 '17 at 14:45
• proofwiki.org/wiki/Matrix_Multiplication_is_Associative – Rust Q Jun 7 '17 at 14:46
• It might be advisable to go through the details, on your own, and post your proof here. When I took abstract for the first time the professor had us do this as an exercise on the very first homework. – Chickenmancer Jun 7 '17 at 14:48
• Associativity of matrix multiplication comes from the identification of square matrices with endomorphisms of $F^n$, and associativity of the composition of maps. It's much shorter this way. – Bernard Jun 7 '17 at 14:51

Matrix multiplication is associative : that's because the multiplicative structure of matrices is a transferred structure from the (algebra) ring of endomorphisms of $K^n$ : since composition is associative, matrix multiplication is as well (note that the formulas of matrix multiplications are, in this set up, theorems rather than definitions)
$I_n$ is an identity element : using the matrix multiplication formulas, this is easily provable. Or else, you can simply note that under the identification $L(K^n)/ M_n(K)$, $I_n$ is the image of the identity mapping $K^n \to K^n$ and therefore it is an identity.
Matrices with nonzero determinant are invertible : This comes from the equalities $M Com(M)^T = det(M)I_n = Com(M)^T M$ where $Com(M)$ is the comatrix of $M$. These equalities hold for matrices over arbitrary rings and can be proved through sheer calculations. Since nonzero scalars are invertible (we are over a field), if $det(M)\neq 0$, this implies $M^{-1} := \frac{1}{det(M)}Com(M)^T$ is an inverse of $M$.
In conclusion, $GL_n(K)$ is a group.