# How is the general linear group a topological group?

How to see if the general linear group GL($n$), of non-singular $n$-square matrices over the real (or complex) numbers under matrix multiplication, is a topological group? How to show that matrix multiplication and inversion are continuous mappings?

-

Hint: Matrix multiplication, entry wise, is just a polynomial! Inversion can be seen to be continuous using the same logic and Cramer's theorem: $\displaystyle A^{-1}=\frac{1}{\det(A)}\text{ adj}(A)$.
Dear Alex, I was having trouble seeing how matrix multiplication is a polynomial, even in the $2 \times 2$ case. I was wondering if you could clarify that point. – user135520 Aug 12 '15 at 19:09
@user135520 It's a polynomial in each entry of the matrix! We can identify an $n \times m$ matrix with $\mathbb{R}^{nm}$, and use the fact that a mapping is continuous iff it is continuous when composed with all of its projection mappings! – Anthony Peter Sep 25 '15 at 6:22
Along with Alex's answer I think this might help you too. Try to think of these,$\phi: M_n(\mathbb{R})\rightarrow \mathbb{R}$ we define $\phi(A)=\det(A)$ is continuous map because $1.$ Polynomials are continuous $2.$ Sum of Two Continuous function is again a continuous. $3.$ Product of two continuous function is again a continuous function.
Could you please explain how $\phi$ is continuous? If we take an open subset of $\mathbb{R}$, why must the pre-image under $\phi$ be open? Is there a topology on $M_n(\mathbb{R})$ a priori? Or is there another way of thinking about continuity in this situation? – Kevin Sheng Oct 17 '15 at 0:46