TITLE: General linear group and special linear group
QUESTION [5 upvotes]: Consider the general linear group $$GL(n,\mathbb R)=\{g\in {\mathbb R}^{n\times n}\mid\det(g)\neq 0\}$$ Prove that the derivative of the function $f=\det:{\mathbb R}^{n\times n}\to\mathbb R$ is given by $$df(g)v=\det(g)\operatorname{trace}(g^{-1}v)$$ for every $g\in GL(n,\mathbb R)$ and every $v\in {\mathbb R}^{n\times n}$. Deduce that the special linear group $$SL(n,\mathbb R):=\{g\in GL(n,\mathbb R)\mid\det(g)=1\}$$ is a smooth submainfold of ${\mathbb R}^{n\times n}$

I have a bunch of questions to ask before I can solve it: Is the dimension of $GL(n,\mathbb R)$ $n^2$? And what is the dimension of the special linear group? What are the basis vectors of the special linear group? Is the general linear space is a metric space? I was trying to compute $df(g)v$, and by definition, $df(g)v=\lim_{t\to 0}\dfrac{\det(g+tv)-\det(g)}{t}$. But how can I expand $
\det(g+tv)$?

REPLY [4 votes]: Some answers/hints:

Yes the dimension of $GL(n,\mathbb{R})$ is $n^2$. In fact it's the open subset of $\mathbb{R}^{n^2}$ on which the determinant (which is continuous) is nonzero.
The special linear group is the subset of the general linear group on which the determinant is one. Neither of these is itself a vector space, so talking about basis vectors doesn't make sense. Here's a hint: If we have $f: M\rightarrow \mathbb{R}$ and consider $N = f^{-1}(\{c\})$, for some $c\in \mathbb{R}$, when is $N$ a manifold? What is its dimension when it is a manifold?
There are various ways to find the derivative of $\det$. Try writing it out in coordinates and using multilinearity of the determinant. (This is easiest when $g=I$, so start there. If you like you can deduce the whole formula from its value at $g=I$ by using the matrix multiplication map $GL_n \times GL_n \rightarrow GL_n$ and the chain rule.)